Câu 1 :

\(a,5\left(x+2\right)=2\left(x-4\right)\)

\(\Leftrightarrow5x+10=2x-8\)

\(\Leftrightarrow5x-2x=-8-10\)

\(\Leftrightarrow3x=-18\)

\(\Leftrightarrow x=-6\)

\(b,x\left(x+2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{3;2\right\}\)

\(c,\dfrac{2x-5}{4}-\dfrac{x+1}{3}=\dfrac{1}{2}\)

\(\Leftrightarrow3\left(2x-5\right)-4\left(x+1\right)=6\)

\(\Leftrightarrow6x-15-4x-4=6\)

\(\Leftrightarrow6x-4x=6+4+15\)

\(\Leftrightarrow2x=25\)

\(\Leftrightarrow x=\dfrac{25}{2}\)

Vậy \(S=\left\{\dfrac{25}{2}\right\}\)

\(d,\dfrac{3}{x-2}-\dfrac{6}{x+2}=\dfrac{-x}{x^2-4}\left(đkxđ:x\ne\pm2\right)\)

\(\Leftrightarrow3\left(x+2\right)-6\left(x-2\right)=-x\)

\(\Leftrightarrow3x+6-6x+12=-x\)

\(\Leftrightarrow3x-6x+x=-12-6\)

\(\Leftrightarrow-2x=-18\)

\(\Leftrightarrow x=9\left(nhận\right)\)

Vậy \(S=\left\{9\right\}\)

Câu 3 : 

a, Xét ΔABD và ΔHBA có :

\(\widehat{A}=\widehat{H}=90^0\)

\(\widehat{B}:chung\)

\(\Rightarrow\Delta ABD\sim\Delta HBA\left(g-g\right)\)

b, Xét ΔADH và ΔDBC có :

\(\widehat{H}=\widehat{C}=90^0\)

\(\widehat{ADH}=\widehat{DBC}\left(AB//CD,slt\right)\)

\(\Rightarrow\Delta ADH\sim\Delta DBC\)

c, Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)

\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)

\(\Rightarrow AB^2=BH.BD\)

d, Xét ΔABD vuông ở A , theo định lý Pi-ta-go ta được :

\(\Rightarrow BD=\sqrt{AB^2+AD^2}=\sqrt{12^2+9^2}=15\left(cm\right)\)

Ta có : \(\Delta ABD\sim\Delta HBA\left(cmt\right)\)

\(\Rightarrow\dfrac{AB}{BH}=\dfrac{BD}{AB}\)

hay \(\dfrac{12}{BH}=\dfrac{15}{12}\)

\(\Rightarrow BH=\dfrac{12.12}{15}=9,6\left(cm\right)\)

Câu 16: A

\(n_{CuSO_4}=0,2x\left(mol\right)\)

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\)

0,2x    0,2x                         0,2x

\(m_{tăng}=m_{Cu}-m_{Fe}=64\cdot0,2x-56\cdot0,2x=1,6\)

\(\Rightarrow x=1M\)

Chọn C.

\(n_{Fe}=n_{FeSO_4}=n_{Cu}=n_{CuSO_4}=0,2.x\left(mol\right)\\ Fe+CuSO_4\rightarrow FeSO_4+Cu\\ m_{t\text{ăn}g}=m_{Cu.b\text{á}m.v\text{ào}}-m_{Fe.tan.ra}\\ \Leftrightarrow1,6=64.0,2x-56.0,2x\\ \Leftrightarrow x=1\\ \Rightarrow C\)

ĐK: \(-1\le x\le1\)

Đặt \(\sqrt{1-x}=a;\sqrt{x+1}=b\Rightarrow3-x=2a^2+b^2\)

\(pt\Leftrightarrow2a-b+3ab=2a^2+b^2\)

\(\Leftrightarrow2a^2+b^2-2a+b-3ab=0\)

\(\Leftrightarrow2a^2-a\left(3b+2\right)+b^2+b=0\)

\(\Delta=\left(3b+2\right)^2-4.2.\left(b^2+b\right)=9b^2+12b+4-8b^2-8b\)

\(=b^2+4b+4=\left(b+2\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}a=\dfrac{3b+2-\left(b+2\right)}{4}=\dfrac{2b}{4}=\dfrac{b}{2}\Leftrightarrow2a=b\left(1\right)\\a=\dfrac{3b+2+b+2}{4}=\dfrac{4b+4}{4}=b+1\left(2\right)\end{matrix}\right.\)

pt (1) \(\Leftrightarrow2\sqrt{1-x}=\sqrt{x+1}\)

\(\Leftrightarrow4\left(1-x\right)=x+1\)

\(\Leftrightarrow5x=3\Leftrightarrow x=\dfrac{5}{3}\left(tm\right)\)

\(pt\left(2\right)\Leftrightarrow\sqrt{1-x}=1+\sqrt{x+1}\)

\(\Leftrightarrow1-x=1+x+1+2\sqrt{x+1}\)

\(\Leftrightarrow-1-2x=2\sqrt{x+1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2+4x+1=4x+4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\4x^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{3}}{2}\left(l\right)\\x=-\dfrac{\sqrt{3}}{2}\left(tm\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy, pt có tập nghiệm là: \(S=\left\{-\dfrac{\sqrt{3}}{2};\dfrac{5}{3}\right\}\)

???

`a,` ĐKXĐ: `x>=0;x\ne1`

`A=...=(sqrtx(1+sqrtx)+sqrtx(1-sqrtx)+sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=(sqrtx+x+sqrtx-x+sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=(3sqrtx-3)/((1-sqrtx)(1+sqrtx))`

`=-3/(1+sqrtx)`

`b,A=-3/(1+sqrtx)` 

Vì `x>=0` nên `1+sqrtx>=1` nên `3/(1+sqrtx)<=3` suy ra `A>=-3`

Dấu "=" xảy ra `<=>x=0`

Vậy `A_(min)=-3<=>x=0`

`a)->` ĐKXĐ : `x>=0;x\ne1`

`b)` Ta có :

`A=((\sqrtx)/(1-\sqrtx)+(\sqrtx)/(1+\sqrtx))+(3-\sqrtx)/(x-1)`

`A=[(\sqrtx(1+\sqrtx)+\sqrtx(1-\sqrtx))/(1-x)]+(3-\sqrtx)/(x-1)`

`A=[(\sqrtx+x+\sqrtx-x)/(1-x)]+(3-\sqrtx)/(x-1)`

`A=(2\sqrtx)/(1-x)+(3-\sqrtx)/(x-1)`

`A=(3-\sqrtx)/(x-1)-(2\sqrtx)/(x-1)`

`A=(3+\sqrtx)/(x-1)`

Vậy `A=(3+\sqrtx)/(x-1)` khi `x>=0;x\ne1`

13 I wish I could help her with her business

14 The children said that they were waiting for the school bus

15 The teacher asked his students to listen to her and not to make a noise

16 My sister studied hard so she completed her exam successfully

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18 She was tired so she went home yesterday

19 He has studied English for four years