a)    4(x + 3)(3x - 2) - 3(x - 1)(4x - 1) = -27
<=> 4(3x² + 7x - 6) - 3(4x² - 5x + 1) = -27
<=> 12x² + 28x - 24 - 12x² + 15x - 3 = -27
<=> 43x = 0 <=> x = 0
Vậy nghiệm là x = 0
b) Đề không rõ, mình sửa lại đề nha:
       4x(2x² - 1) + 27 = (4x² + 6x + 9)(2x + 3)
<=> 8x³ - 4x + 27 = 8x³ + 24x² + 36x + 27
<=> 24x² + 40x = 0 <=> x = 0 hay x = -5/3
Vậy nghiệm là x = 0 hay x = -5/3
 

2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

Bài giải

\(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)

\(8x^3-4x+27=8x^3+12x^2+18x+12x^2+18x+27\)

\(8x^3-4x+27=8x^3+24x^2+36x+27\)

\(8x^3-4x+27-8x^3-36x-27=24x^2\)

\(-40x=24x^2\)

\(\frac{3}{5}x^2=x\)

\(\frac{3}{5}x^2-x=0\)

\(x\left(\frac{3}{5}x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\frac{3}{5}x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{5}{3}\end{matrix}\right.\)

\(\Rightarrow\text{ }x\in\left\{0\text{ ; }\frac{5}{3}\right\}\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

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

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

1/

\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Rightarrow x=0\)

2/

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)

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

\(\Leftrightarrow x=0\)

1.A =( x-3)( x+3) + 15 - x²

   A=X²-3X+3X+15-X³

  A=15-X

2.B=(X -1) (X²+X+1) - X (X²+2) + 2X  

 B=X³+ X²+ X - X² - X - 1 - X³ - 2X + 2X

B=   -1

3.C=(2X - 1 ) (4X² + 2X + 1) - X ( 8 X 2 + 1 ) + X

C=8X³ - 4X² +4X² - 2X +2 X - 1 - 8X²2 - X + X

C=8X³ - 1 - 8X²2

MK CHỈ LM ĐC TỚI ĐÓ THUI SAI CHỖ NÀO ĐỪNG TRÁCH VÌ MK YẾU PHẦN NÀY

Ta có: \(4x\left(2x^2-1\right)+27=\left(4x^2+6x+9\right)\left(2x+3\right)\)

\(\Leftrightarrow8x^3-4x+27=8x^3+12x^2+12x^2+18x+18x+27\)

\(\Leftrightarrow8x^3-4x+27-8x^3-24x^2-36x-27=0\)

\(\Leftrightarrow-24x^2-40x=0\)

\(\Leftrightarrow-8x\left(3x+5\right)=0\)

mà -8≠0

nên \(\left[{}\begin{matrix}x=0\\3x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-5}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{0;\frac{-5}{3}\right\}\)