a)

ĐKXĐ: \(x> \frac{-5}{7}\)

Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)

\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)

\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)

Vậy......

b) ĐKXĐ: \(x\geq 5\)

\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}\sqrt{9}.\sqrt{x-5}=4\)

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)

(hoàn toàn thỏa mãn)

Vậy..........

c) ĐK: \(x\in \mathbb{R}\)

Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)

\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

Khi đó:

\(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)

\(\Leftrightarrow 7-a^2+6a=0\)

\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)

\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)

\(\Rightarrow 6x^2-12x+7=a^2=49\)

\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)

(đều thỏa mãn)

Vậy..........

\(C=\dfrac{4x-\left|3x-1\right|}{1-49x^2}\)

Vì x<1/3 nên 3x-1<0

=>\(C=\dfrac{4x-\left(1-3x\right)}{1-49x^2}=\dfrac{7x-1}{-\left(7x-1\right)\left(7x+1\right)}=\dfrac{-1}{7x+1}\)

\(A=3x+\sqrt{9x^2-24x+16}=3x+\sqrt{\left(3x\right)^2-2.3.4x+4^2}=3x+\sqrt{\left(3x-4\right)^2}=3x+\left|3x-4\right|=-9+13=4\)

\(B=5x-\sqrt{\left(2x\right)^2+2.2.3x+3^2}=5x-\sqrt{\left(2x+3\right)^2}=5x-\left|2x+3\right|=-5\sqrt{5}+2\sqrt{5}-3=-3\left(\sqrt{5}+1\right)\)

Giúp e với ạ T.T

2, \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)

\(\Leftrightarrow-\sqrt{x+1}=-17\)

\(\Leftrightarrow x+1=289\left(x>0\right)\)

\(\Leftrightarrow x=288\)

Vậy x = 288

3, \(-5x+7\sqrt{x}+12=0\)

\(\Leftrightarrow-5x+12\sqrt{x}-5\sqrt{x}+12=0\)

\(\Leftrightarrow\sqrt{x}\left(12-5\sqrt{x}\right)+\left(12-5\sqrt{x}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(12-5\sqrt{x}\right)=0\)

Do \(\sqrt{x}+1>0\)

\(\Rightarrow12-5\sqrt{x}=0\Leftrightarrow x=\dfrac{144}{25}\)

Vậy...

1. (Đề có chút sai sai nên mình sửa lại nhé) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)

(ĐK: \(x\ge1\))

\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)

\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)

\(\Leftrightarrow2\sqrt{x-1}=16\)

\(\Leftrightarrow\sqrt{x-1}=8\)

\(\Leftrightarrow x-1=64\)

\(\Leftrightarrow x=65\left(tm\right)\)

Vậy pt đã cho có nghiệm x=65.

2. \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)

(ĐK: \(x\ge-1\))

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9\left(x+1\right)}+3\sqrt{x+1}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)

\(\Leftrightarrow-\sqrt{x+1}=-17\)

\(\Leftrightarrow\sqrt{x+1}=17\)

\(\Leftrightarrow x+1=289\)

\(\Leftrightarrow x=288\left(tm\right)\)

Vậy \(S=\left\{288\right\}\)

3. \(-5x+7\sqrt{x}+12=0\) (ĐK: \(x\ge0\))

\(\Leftrightarrow5x-7\sqrt{x}-12=0\)

\(\Leftrightarrow5x+5\sqrt{x}-12\sqrt{x}-12=0\)

\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\5\sqrt{x}-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vô.lý\right)\\5\sqrt{x}=12\end{matrix}\right.\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\)

Vậy pt có nghiệm \(x=\dfrac{144}{25}\)

a) \(\Leftrightarrow\sqrt{\left(x+3\right)^2}=4\)

\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )

b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)

\(\Leftrightarrow\left|2x-1\right|=5x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)

\(\Leftrightarrow x+3=4\)

\(\Rightarrow x=1\)

a,

\(\sqrt{1-4x+4x^2}=5\\ \sqrt{\left(2x-1\right)^2}=5\\ \left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\\ \left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

b,

\(\sqrt{4-5x}=12\\ 4-5x=144\\ x=-28\)

a) CĂN ký hiệu =v nhé

8 = 2.2² ; x² -4xy + (2y)² = (x-2y)²

=> A = 2v2/(x-2y)

b;c tương tự

+) ta có : \(A=6x+\sqrt{9x^2-12x+4}=6x+\sqrt{\left(3x-2\right)^2}\)

\(=6x+\left|3x-2\right|\) \(\Rightarrow\left[{}\begin{matrix}A=9x-2\left(x\ge\dfrac{2}{3}\right)\\A=3x+2\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)

+) ta có : \(B=5x-\sqrt{x^2+4x+4}=5x-\sqrt{\left(x+2\right)^2}\)

\(=5x-\left|x+2\right|\) \(\Rightarrow\left[{}\begin{matrix}A=4x-2\left(x\ge-2\right)\\6x+2\left(x< -2\right)\end{matrix}\right.\)