\(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\)ĐK : x>= 0 

\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)

\(\Leftrightarrow-11\sqrt{2x}+6x+6=6x-5\Leftrightarrow-11\sqrt{2x}=-11\)

\(\Leftrightarrow\sqrt{2x}=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

a)\(2x^4-6x^3+x^2+6x-3=0\)

\(\Leftrightarrow2x^4-6x^3+3x^2-2x^2+6x-3=0\)

\(\Leftrightarrow x^2\left(2x^2-6x+3\right)-\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x^2-1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x^2-6x+3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x+1=0\\2x^2-6x+3=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\\Delta_{2x^2-6x+3}=\left(-6\right)^2-4\left(2.3\right)=12\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=-1\\x_{1,2}=\frac{6\pm\sqrt{12}}{4}\end{array}\right.\)

b)\(x^3+9x^2+26x+24=0\)

\(\Leftrightarrow x^3+5x^2+6x+4x^2+20x+24=0\)

\(\Leftrightarrow x\left(x^2+5x+6\right)+4\left(x^2+5x+6\right)=0\)

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+2=0\\x+3=0\\x+4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=-3\\x=-4\end{array}\right.\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=18:18\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

b) \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x-5\right)^2=x^2+6x+64\)

\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2-\left(x^2+6x+64\right)=0\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)

\(\Leftrightarrow8^2-x^2-6x-64=0\)

\(\Leftrightarrow64-x^2-6x-64=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow x\left(-x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=-6\)

a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)

\(\Leftrightarrow\left(6x^2+21x-2x-7\right)-\left(6x^2-5x+6x-5\right)-16=0\)

\(\Leftrightarrow6x^2+21x-2x-7-6x^2+5x-6x+5-16=0\)

\(\Leftrightarrow18x-18=0\)

\(\Leftrightarrow18x=18\)

\(\Leftrightarrow x=18:18\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

b, \(\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-5\right)+\left(2x- 5\right)^2=x^2+6x+64\)

\(\Leftrightarrow\left[\left(2x+3\right)-\left(2x-5\right)\right]^2- \left(x^2+6x+64\right)=0\)

\(\Leftrightarrow\left(2x+3-2x+5\right)^2-x^2-6x-64=0\)

\(\Leftrightarrow8^2-x^2-6x-64=0\)

\(\Leftrightarrow64-x^2-6x-64=0\)

\(\Leftrightarrow-x^2-6x=0\)

\(\Leftrightarrow x\left(-x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=6\)

a) (3x−1)(2x+7)−(x+1)(6x−5)=16(3x−1)(2x+7)−(x+1)(6x−5)=16

⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0⇔(6x2+21x−2x−7)−(6x2−5x+6x−5)−16=0

⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0⇔6x2+21x−2x−7−6x2+5x−6x+5−16=0

⇔18x−18=0⇔18x−18=0

⇔18x=18⇔18x=18

⇔x=18:18⇔x=18:18

⇔x=1⇔x=1

Vậy x=1x=1

b) (2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64(2x+3)2−2(2x+3)(2x−5)+(2x−5)2=x2+6x+64

⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0⇔[(2x+3)−(2x−5)]2−(x2+6x+64)=0

⇔(2x+3−2x+5)2−x2−6x−64=0⇔(2x+3−2x+5)2−x2−6x−64=0

⇔82−x2−6x−64=0⇔82−x2−6x−64=0

⇔64−x2−6x−64=0⇔64−x2−6x−64=0

⇔−x2−6x=0⇔−x2−6x=0

⇔x(−x−6)=0⇔x(−x−6)=0

⇔[x=0−x−6=0⇔[x=0−x−6=0

⇔[x=0−x=6⇔[x=0−x=6

⇔[x=0x=−6⇔[x=0x=−6

Vậy x=0x=0 hoặc x=−6

bạn ơi ở 2(2x²+3)hay là 2(2x+3)

(2x+3)^2-2(2x+3)(2x-5)+(2x-5)^2=x^2+6x+64

=>x^2+6x+64=(2x+3-2x+5)^2

=>x^2+6x+64=64

=>x(x+6)=0

=>x=0 hoặc x=-6

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^

⇔(2 x ) ²+2.2 x .1+1 ²=0 ... c ,(3 x −4) ²−14(3 x −4)(6+3 x )+49(3 x +6)=16 ... ⇔9 x ²−24 x +16−126 x ²−252 x +168 x +336+147 x +294=16.

https://olm.vn/hoi-dap/detail/192758180810.html

\(-2x\left(6x-2\right)+3x\left(4x-7\right)=8\)

<=> \(-12x^2+4x+12x^2-21x=8\)

<=> \(-17x=8\)

<=> \(x=-\frac{8}{17}\)

\(\left(7x-2\right)\left(2x+3\right)-\left(3x-5\right)\left(4x+6\right)=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30=2x^2-5\)

<=> \(14x^2+17x-6-12x^2+2x+30-2x^2+5=0\)

<=> \(19x+29=0\)

<=> \(19x=-29\)

<=> \(x=-\frac{29}{19}\)

Ý cuối mình k biết -22x2 là -22.2 hay -22x² nữa :)