Đề thi HK1 Toán 8 - Trường Tạ Quang Bửu - Hà Nội - Năm 2017 - 2018 (có giải chi tiết).

Câu 1 : Thực hiện các phép tính:1. \(2xy\left( {x + 2y} \right)\)2. \(\left( {x + 1} \right)\left( {2x - 1} \right)\)3. \(10{x^4}{y^3}:6{x^2}{y^2}\)4. \(\left( {{x^3} - 8} \right):\left( {{x^2} + 2x + 4} \right)\)

A \(\begin{array}{l}1.\,\,2{x^2}y + 2x{y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.\,\,\frac{5}{3}{x^2}y\\2.\,\,2{x^2} + x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4.\,\,x - 2\end{array}\)

B \(\begin{array}{l}1.\,\,2{x^2}y + 2x{y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.\,\,\frac{5}{3}xy\\2.\,\,2{x^2} + x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4.\,\,x + 2\end{array}\)

C \(\begin{array}{l}1.\,\,2{x^2}y + 2x{y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.\,\,\frac{5}{3}{x^2}y\\2.\,\,2{x^2} - x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4.\,\,x + 2\end{array}\)

D \(\begin{array}{l}1.\,\,2{x^2}y + 2x{y^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3.\,\,\frac{5}{3}{x^2}y\\2.\,\,2{x^2} - x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4.\,\,x - 2\end{array}\)

Câu 2 : Phân tích đa thức thành nhân tử:1) \(2x{y^2} - 4y\)2) \({x^2}y - 6xy + 9y\)3) \({x^2} + x - {y^2} + y\)4) \({x^2} + 4x + 3\)

A \(\begin{array}{l}1)\,\,2y\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x - 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

B \(\begin{array}{l}1)\,\,2{y^2}\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x + 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

C \(\begin{array}{l}1)\,\,2y\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x - y} \right)\left( {x + y + 1} \right)\\2)\,\,y{\left( {x + 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x + 3} \right)\left( {x + 1} \right)\end{array}\)

D \(\begin{array}{l}1)\,\,2{y^2}\left( {xy - 2} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,\left( {x + y} \right)\left( {x - y + 1} \right)\\2)\,\,y{\left( {x - 3} \right)^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,\left( {x - 3} \right)\left( {x - 1} \right)\end{array}\)

Câu 3 : Cho biểu thức: \(P = \frac{{2{x^2} - 1}}{{{x^2} + x}} - \frac{{x - 1}}{x} + \frac{3}{{x + 1}}\)1. Rút gọn \(P\) .2. Tìm x để \(P = 0\)3. Tính giá trị biểu thức \(P\) khi \(x\) thỏa mãn: \({x^2} - x = 0\).4. Tìm giá trị lớn nhất của biểu thức \(Q = \frac{1}{{{x^2} - 9}}.P\)

A \(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)

B \(\begin{array}{l}1)\,\,P = \frac{{x + 3}}{{x + 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = 2\\2)\,\,x = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)

C \(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = - 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{{ - 1}}{4}\end{array}\)

D \(\begin{array}{l}1)\,\,P = \frac{{x - 3}}{{x - 1}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3)\,\,P = - 2\\2)\,\,x = - 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4)\,\,maxQ = \frac{1}{4}\end{array}\)

Câu 4 : Cho \(\Delta ABC\) vuông tại \(A,\,AB = 6\,cm,\,AC = 8\,cm\). Gọi \(M\) là trung điểm của đoạn \(BC\) . Điểm \(D\) đối xứng với \(A\) qua \(M\) .1. Chứng minh tứ giác \(AB{\rm{D}}C\) là hình chữ nhật. Tính diện tích hình chữ nhật \(AB{\rm{D}}C\) .2. Kẻ \(AH \bot BC\left( {H \in BC} \right)\), gọi \(E\) là điểm đối xứng với \(A\) qua \(H\) . Chứng minh: \(HM//DE\) và \(HM = \frac{1}{2}DE\).3. Tính tỉ số \(\frac{{{S_{AHM}}}}{{{S_{A{\rm{ED}}}}}}\).4. Chứng minh tứ giác \(BC{\rm{D}}E\) là hình thang cân.

Đề thi HK1 Toán 8 - Trường Tạ Quang Bửu - Hà Nội -...