Giải chi tiết bài lm ra nha m.n ;-;

1. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{x-1}=13-x$

\(\Rightarrow \left\{\begin{matrix} 13-x\geq 0\\ x-1=(13-x)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ x^2-27x+170=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 13\\ (x-17)(x-10)=0\end{matrix}\right.\)

\(\Rightarrow x=10\) (tm)

2. ĐKXĐ: $x\geq 3$

\(3\sqrt{x+34}-3\sqrt{x-3}=1\)

\(\Leftrightarrow 3\sqrt{x+34}=3\sqrt{x-3}+1\)

\(\Rightarrow 9(x+34)=9x+6\sqrt{x-3}-26\)

\(\Leftrightarrow \frac{166}{3}=\sqrt{x-3}\)

$\Leftrightarrow x-3=\frac{27556}{9}$

$\Leftrightarrow x=\frac{27583}{9}$ (tm)

3.

ĐKXĐ: $x\ge \frac{5}{3}$

\(\sqrt{2x+5}-\sqrt{3x-5}=2\)

$\Leftrightarrow \sqrt{2x+5}=2+\sqrt{3x-5}$

$\Leftrightarrow 2x+5=3x+4\sqrt{3x-5}-1$

$\Leftrightarrow 6-x=4\sqrt{3x-5}$
\(\Rightarrow \left\{\begin{matrix} 6-x\geq 0\\ (6-x)^2=16(3x-5)\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq 6\\ x^2-60x+116=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\leq 6\\ (x-58)(x-2)=0\end{matrix}\right.\Rightarrow x=2\)

Thử lại thấy tm

`P=(1+5/(sqrtx-2)).(sqrtx-(x+2sqrtx+4)/(sqrtx+3))`

`=((sqrtx-2+5)/(sqrtx-2)).((x+3sqrtx-x-2sqrtx-4)/(sqrtx+3))`

`=(sqrtx+3)/(sqrtx-2).(sqrtx-4)/(sqrtx+3)`

`=(sqrtx-4)/(sqrtx-2)`

Ta có: \(P=\left(1+\dfrac{5}{\sqrt{x}-2}\right)\left(\sqrt{x}-\dfrac{x+2\sqrt{x}+4}{\sqrt{x}+3}\right)\)

\(=\dfrac{\sqrt{x}-2+5}{\sqrt{x}-2}\cdot\dfrac{x+3\sqrt{x}-x-2\sqrt{x}-4}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

Bài 7:

Ta có: \(P=\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+\sqrt{x}-x-1}\right):\left(1+\dfrac{\sqrt{x}}{x+1}\right)\)

\(=\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)

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

Mọi người chỉ cần nói là chọn đáp án nào thôi ạ, không cần phải ghi lời giải quá chi tiết, cảm ơn mọi người trước!

Câu 4:

Gọi 2 cạnh là a,b(cm)(a,b>0)

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{70:2}{5}=7\)

\(\Rightarrow\left\{{}\begin{matrix}a=7.2=14\left(cm\right)\\b=7.3=21\left(cm\right)\end{matrix}\right.\)

\(\Rightarrow S=a.b=14.21=294\left(cm^2\right)\)

Câu 9:

\(2x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{x+y}{5+2}=\dfrac{-42}{7}=-6\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-6\right).5=-30\\y=\left(-6\right).2=-12\end{matrix}\right.\)

Câu 10:

C

4a

9a

10c