https://hoc24.vn/cau-hoi/a-5x-2x-62-50b-5x-x-150-2-3c-6x-x-511-59-31d-5x-3x-36-334-124x-2x-68-219-216.2785429565572

a: \(\Leftrightarrow7x=35\)

hay x=5

b: \(\Leftrightarrow6x=78\)

hay x=13

a) \(x\in\left\{15;30;45\right\}\)

b) \(x\in\left\{12;24;36\right\}\)

c) \(x\in\left\{60;120\right\}\)

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)

ở oooo

e: \(\Leftrightarrow3x\left(x-4\right)=0\)

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

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1: \(\Leftrightarrow x=UCLN\left(24;36;150\right)=6\)

2: \(\Leftrightarrow x\in\left\{24;48;72;...\right\}\)

mà 16<=x<=50

nên \(x\in\left\{24;48\right\}\)

3: \(\Leftrightarrow x\inƯ\left(6\right)\)

mà x>-10

nên \(x\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

4: \(\Leftrightarrow x\in BC\left(4;5;8\right)\)

\(\Leftrightarrow x\in\left\{...;-40;0;40;80;120;160;200;...\right\}\)

mà -20<x<180

nên \(x\in\left\{0;40;80;120;160\right\}\)

a: Ta có: \(\sqrt{\sqrt{x}+3}=4\)

\(\Leftrightarrow\sqrt{x}+3=16\)

\(\Leftrightarrow\sqrt{x}=13\)

hay x=169

b: Ta có: \(\sqrt{x+3}=\sqrt{1-5x}\)

\(\Leftrightarrow x+3=1-5x\)

\(\Leftrightarrow6x=-2\)

hay \(x=-\dfrac{1}{3}\left(nhận\right)\)

a) \(\sqrt{3+\sqrt{x}}=4\left(đk:x\ge0\right)\)

\(\Leftrightarrow3+\sqrt{x}=16\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\left(tm\right)\)

b) \(\sqrt{x+3}=\sqrt{1-5x}\left(đk:\dfrac{1}{5}\ge x\ge-3\right)\)

\(\Leftrightarrow x+3=1-5x\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\left(ktm\right)\)

Vậy \(S=\varnothing\)

c) \(\sqrt{x^2+6x+9}=3x-1\left(đk:x\ge\dfrac{1}{3}\right)\)

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

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

\(\Leftrightarrow x+3=3x-1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

a. \(\sqrt{3+\sqrt{x}}=4\)      ĐKXĐ: \(x\ge0\)

<=> 3 + \(\sqrt{x}\) = 4²

<=> \(3+\sqrt{x}=16\)

<=> \(\sqrt{x}=16-3\)

<=> \(\sqrt{x}=13\)

<=> x = 13²

<=> x = 169 (TM)

b. \(\sqrt{x+3}=\sqrt{1-5x}\)           ĐKXĐ: \(x\ge\dfrac{1}{5}\)

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

<=> \(|x+3|=|1-5x|\)

<=> \(\left[{}\begin{matrix}x+3=1-5x\\-\left(x+3\right)=-\left(1-5x\right)\\x+3=-\left(1-5x\right)\\-\left(x+3\right)=1-5x\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-1}{3}\\x=1\\x=1\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=1\end{matrix}\right.\)

c. \(\sqrt{x^2+6x+9}=3x-1\)

<=> \(\sqrt{\left(x+3\right)^2}=3x-1\)

<=> \(|x+3|=3x-1\)

<=> \(\left[{}\begin{matrix}x+3=-\left(3x-1\right)\\x+3=3x-1\end{matrix}\right.\)

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

<=> \(\left[{}\begin{matrix}4x=-2\\x=2\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=2\end{matrix}\right.\)

a, \(\left(2x-1\right)\left(x+3\right)-2x^2+5x=7\)

\(\Leftrightarrow2x^2+6x-x-3-2x^2+5x=7\)

\(\Leftrightarrow2x^2+5x-3-2x^2+5x=7\)

\(\Leftrightarrow10x-10=0\Leftrightarrow x=1\)

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-4\right)\left(x+4\right)=54\)

\(\Leftrightarrow\left(x^3+27\right)-x\left(x^2-16\right)=54\)

\(\Leftrightarrow x^3+27-x^3+16x=54\)

\(\Leftrightarrow-27+16x=0\Leftrightarrow x=\frac{27}{16}\)

a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

c: Ta có: \(\left(x-3\right)^2-4=0\)

\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)

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

b. 

PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$

$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$

$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$

$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$

$\Leftrightarrow (x-3)^2(2x+1)^2=0$

$\Leftrightarrow (x-3)(2x+1)=0$

$\Leftrightarrow x-3=0$ hoặc $2x+1=0$

$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$

d.

$x^2-2x=24$

$\Leftrightarrow x^2-2x-24=0$

$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$

$\Leftrightarrow x=-4$ hoặc $x=6$