:) chán thật sự 

lm câu 14;16;18;20;22 hả

14)3x-2=2x-3

=>3x-2-2x+3=0

=>x-1=0

=>x=1

16)\(\dfrac{5x+3}{12}\)=\(\dfrac{1+2x}{9}\)

=>\(\dfrac{\left(5x+3\right)x3}{12x3}\)=\(\dfrac{\left(1+2x\right)x3}{9x3}\)

Khử mẫu

=>5x⁴+3x³=x³+2x⁴

=>5x⁴+3x³-x³+2x⁴=0

=>7x⁴+2x³=0

=>x³(7x+2)=0

=>x=0 hoặc 7x+2=0=>x=-2/7

Bài 1: 

ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

a) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{\left(x+2\right)\left(x-2\right)}{x+2}+\dfrac{10-x^2}{x+2}\right)\)

\(=\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{6}\)

\(=\dfrac{-1}{x-2}\)

b) Ta có: \(\left|x\right|=\dfrac{1}{2}\)

nên \(x\in\left\{\dfrac{1}{2};\dfrac{-1}{2}\right\}\)

Thay \(x=\dfrac{1}{2}\) vào biểu thức \(A=\dfrac{-1}{x-2}\), ta được:

\(A=-1:\left(\dfrac{1}{2}-2\right)=-1:\dfrac{-3}{2}=\dfrac{-1\cdot2}{-3}=\dfrac{2}{3}\)

Thay \(x=-\dfrac{1}{2}\) vào biểu thức \(A=\dfrac{-1}{x-2}\), ta được:

\(A=-1:\left(-\dfrac{1}{2}-2\right)=-1:\dfrac{-5}{2}=1\cdot\dfrac{2}{5}=\dfrac{2}{5}\)

Vậy: Khi \(\left|x\right|=\dfrac{1}{2}\) thì \(A\in\left\{\dfrac{2}{3};\dfrac{2}{5}\right\}\)

c) Để A<0 thì \(\dfrac{-1}{x-2}< 0\)

\(\Leftrightarrow x-2>0\)

hay x>2

Kết hợp ĐKXĐ, ta được: x>2

Vậy: Để A<0 thì x>2

2x:1=-7/2

adu Huyền lên mạng hỏi bài, t lên mạng tra bài :]]]]

23D

24A

25C

26C

27D

29D

câu 12:

Sử dụng quỹ tích điểm:

\(T_{\overrightarrow{V}}\left(M\right)=M'\left(X';Y'\right)\) với mọi điểm \(M\left(x';y'\right)\in\left(E\right)\Rightarrow\left\{{}\begin{matrix}X=X'-2\\Y=Y'-2\end{matrix}\right.\)

\(\left(E\right):\dfrac{\left(x-2\right)^2}{16}+\dfrac{\left(y-1\right)^2}{9}=1\)

câu 13;

\(\text{✳}d=MN=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

\(\text{✳}d'=M'N'=\sqrt{\left(X'_2-X'_1\right)^2+\left(Y'_2-Y'_1\right)^2}\)

✱\(\left(X'_2-X'_1\right)^2\)

\(=\left[\left(X_2-X_1\right)\cos a+\left(-Y_2+Y_1\right)\sin a\right]^2\)

\(=\left[\left(X_2-X_1\right)\cos a-\left(Y_2-Y_1\right)\sin a\right]^2\)

\(=\left(X_2-X_1\right)^2\cos^2a+\left(Y_2-Y_1\right)^2\sin^2a-2\left(X_2-X_1\right)\left(Y_2-Y_1\right)\sin a.\cos a\)

✱\(\left(Y'_2-Y'_1\right)^2\)

\(=\left[\left(X_2-X_1\right)\sin a-\left(Y_2-Y_1\right)\cos a\right]^2\)

\(=\left(X_2-X_1\right)^2\sin^2a+\left(Y_2-Y_1\right)^2\cos^2a-2\left(X_2-X_1\right)\left(Y_2-Y_1\right)\sin a.\cos a\)

vậy \(d'=\sqrt{=\left(X_2-X_1\right)^2\left(\cos^2a+\sin^2a\right)+\left(Y_2-Y_1\right)^2\left(\sin^2a+\cos^2a\right)}\)

HAY \(d'=\sqrt{\left(X_2-X_1\right)^2+\left(Y_2-Y_1\right)^2}\) (vì \(\cos^2a+\sin^2a=1\))

Bài 5: 

n_{\(Na_2ZnO_2\)} = \(\dfrac{8,58}{143}=0,06mol\)

n_{\(Mg_3\left(PO_4\right)_2\)} =\(\dfrac{7,86}{262}=0,03mol\)

\(n_{CuSO_4}=\dfrac{7,42}{160}=0,046375\approx0,046mol\)

\(n_{CO_2}=\dfrac{26,88}{22,4}=1,2mol\)

\(n_{NO}=\dfrac{20,16}{22,4}=0,9mol\)

\(n_{NO_2}=\dfrac{16,8}{22,4}=0,75mol\)

\(n_{CuSO_4}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2mol\)

\(n_{Fe}=\dfrac{4,8.10^{23}}{6.10^{23}}=0,8mol\)

n_{\(Na_3PO_4\)} = \(\dfrac{15,3.10^{23}}{6.10^{23}}=2,55mol\)