a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

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

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

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

\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)

\(\Leftrightarrow a=b\)

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

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)

ĐKXĐ: \(x>0\)

\(3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(x+\dfrac{1}{4x}+1\right)-9\)

\(\Leftrightarrow3\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)< 2\left(\sqrt{x}+\dfrac{1}{2\sqrt{x}}\right)^2-9\)

Đặt \(\sqrt{x}+\dfrac{1}{2\sqrt{x}}=a>0\)

\(\Rightarrow3a< 2a^2-9\Rightarrow2a^2-3a-9>0\)

\(\Rightarrow\left(a-3\right)\left(2a+3\right)>0\)

\(\Rightarrow a-3>0\Rightarrow a>3\)

\(\Rightarrow\sqrt{x}+\dfrac{1}{2\sqrt{x}}>3\Leftrightarrow2x+1>6\sqrt{x}\)

\(\Leftrightarrow2x-6\sqrt{x}+1>0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}>\dfrac{3+\sqrt{7}}{2}\\0\le\sqrt{x}< \dfrac{3-\sqrt{7}}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x>\dfrac{8+3\sqrt{7}}{2}\\0\le x< \dfrac{8-3\sqrt{7}}{2}\end{matrix}\right.\)

(x - 1)(2x² - 10) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x^2-10=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2=10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{1;\sqrt{5}\right\}\)
(2x - 7)² - 6(2x - 7)(x - 3) = 0

\(\Leftrightarrow\left(2x-7\right)\left(2x-7-6x+18\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(11-4x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=7\\4x=11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là: \(S=\left\{\frac{7}{2};\frac{11}{4}\right\}\)
(5x + 3)(x² + 4) = 0

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x^2+4=0\end{matrix}\right.\)

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

\(\Leftrightarrow x=-\frac{3}{5}\)

Vậy phương trình có tập nghiệm là: \(S=\left\{-\frac{3}{5}\right\}\)

a)

\(\left(x-1\right)\cdot\left(2x^2-10\right)=0\\ \Leftrightarrow\left(x-1\right)\cdot2\cdot\left(x^2-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-1=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{5}\end{matrix}\right.\)

b)

\(\left(2x-7\right)^2-6\cdot\left(6x-7\right)\cdot\left(x-3\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left[\left(2x-7\right)-6\cdot\left(x-3\right)\right]=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(2x-7-6x+18\right)=0\\ \Leftrightarrow\left(2x-7\right)\cdot\left(11-4x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-7=0\\11-4x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=\frac{11}{4}\end{matrix}\right.\)

c)

\(\left(5x+3\right)\cdot\left(x^2+4\right)=0\)

Vì \(\left(x^2+4\right)>0\Rightarrow\left(loại\right)\)

\(\Rightarrow5x+3=0\\ \Rightarrow x=-\frac{3}{5}\)

mình lười nên nói cách làm nhé

B1: chuyển \(\dfrac{6}{x^2-9}\)sang vế trái và thêm dấu trừ ở trc \(\dfrac{6}{x^2-9}\)và vế phải =0

B2: để ý thấy \(x^2-9\)=(x-3).(x+3) tức là hằng đẳng thức số 3 ý

B3: quy đồng mẫu , mẫu số chung là (x-3).(x+3).(2x+7)

B4: chia cả hai vế cho (x-3).(x+3).(2x+7)

lưu ý : bước này là dấu⇒ chứ ko phải dấu ⇔ nhé

B5: giải pt như bình thg thui

ĐKXĐ: \(x\notin\left\{3;-3;-\dfrac{7}{2}\right\}\)

Ta có: \(\dfrac{13}{\left(x-3\right)\left(2x+7\right)}+\dfrac{1}{2x+7}=\dfrac{6}{x^2-9}\)

\(\Leftrightarrow\dfrac{13\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}+\dfrac{x^2-9}{\left(2x+7\right)\left(x-3\right)\left(x+3\right)}=\dfrac{6\left(2x+7\right)}{\left(x-3\right)\left(x+3\right)\left(2x+7\right)}\)

Suy ra: \(13x+39+x^2-9=12x+42\)

\(\Leftrightarrow x^2+13x+30-12x-42=0\)

\(\Leftrightarrow x^2+x-12=0\)

\(\Leftrightarrow x^2+4x-3x-12=0\)

\(\Leftrightarrow x\left(x+4\right)-3\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-4}

a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)

TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)

TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)

b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)

c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)

\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)

Câu 1: 

a: x+2=0

nên x=-2

b: (x-3)(2x+8)=0

=>x-3=0 hoặc 2x+8=0

=>x=3 hoặc x=-4

a . 

x + 2 = 0

=> x = 0 - 2 = -2 

b ) .

<=> x - 3 = 0 ; 2x + 8 = 0

= > x = 3 ; x = -8/2 = -4 

c ) .

ĐKXĐ của pt : x - 5 khác 0 = > ddk : x khác 5

1)

a) \(x+2=0\)

\(\Leftrightarrow x=-2\)

Vậy S = {\(-2\)}

b) \(\left(x-3\right)\left(2x+8\right)=0\)

\(\Leftrightarrow x-3=0\) hoặc \(2x+8=0\)

*) \(x-3=0\)

\(\Leftrightarrow x=3\)

*) \(2x+8=0\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\)

Vậy S = \(\left\{-4;3\right\}\)

2) ĐKXĐ:

\(x-5\ne0\Leftrightarrow x\ne5\)

ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{2x-4+2\sqrt{2x-5}}+\sqrt{2x+4+6\sqrt{2x-5}}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}+1\right)^2}+\sqrt{\left(\sqrt{2x-5}+3\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{2x-5}+1\right|+\left|\sqrt{2x-3}+3\right|=14\)

\(\Leftrightarrow2\sqrt{2x-5}=10\)

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

\(\Leftrightarrow2x-5=25\)

\(\Leftrightarrow x=15\)

Lời giải:

ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow (3x^2-10x-25)=2(x+3)(\sqrt{2x-1}-3)$

$\Leftrightarrow (x-5)(3x+5)=2(x+3).\frac{2(x-5)}{\sqrt{2x-1}+3}$

\(\Leftrightarrow (x-5)\left[(3x+5)-\frac{4(x+3)}{\sqrt{2x-1}+3}\right]=0\)

Xét biểu thức trong ngoặc vuông:

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

\(\Leftrightarrow (3x+5)\sqrt{2x-1}=-(3+5x)\)

Dễ thấy điều này vô lý vì với $x\geq \frac{1}{2}$ thì vế trái không âm còn vế phải âm.

Vậy $x-5=0\Leftrightarrow x=5$

\(\Leftrightarrow\left(4x+14\right)^2=\left(3x+9\right)^2\)

\(\Leftrightarrow\left(4x+14+3x+9\right)\cdot\left(4x+14-3x-9\right)=0\)

\(\Leftrightarrow\left(7x+23\right)\left(x+5\right)=0\)

hay \(x\in\left\{-\dfrac{23}{7};-5\right\}\)

\(\Leftrightarrow\left(8x+14\right)^2=\left(9x+3\right)^2\)

\(\Leftrightarrow8x+14=9x+3\)

\(\Leftrightarrow x=11\)