Cái này thì cứ nhân từ từ để phá dấu ngoặc rùi tính ra thôi !!!

\(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Leftrightarrow5x-3\left\{4x-2\left[6-11x\right]\right\}=182\)

\(\Leftrightarrow5x-3\left\{26x-12\right\}=182\)

\(\Leftrightarrow5x-78x+36=182\)

\(\Leftrightarrow-73x=146\)

\(\Leftrightarrow x=-2\)

TK NKA !!!

dewrerrrrrrtt

dễ hơn ăn cơm sườn;

=5x -12x+45x -18=182

38x = 200

x= 200/38=100/19

b) \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3\right)+\left(x+2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x+3+x+5\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

chán òi ko lm nữa đâu có ng đg bùn mk cx bùn theo lun xl nha

1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)

\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )

2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)

\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)

\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )

Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))

1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

=> \(-4x^2+28x+4x^3-20x=28x^2-13\)

=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)

=> \(-4x^2+4x^3+8x-28x^2+13=0\)

=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)

=> \(-32x^2+4x^3+8x+13=0\)

=> vô nghiệm

2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)

=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)

=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)

=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)

=> \(-14x^2-56x+12=0\)

=> .... tự tìm

Câu c dấu bằng chỗ nào ?

\(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)

\(< =>-4x^2+28x+4x^3-20x=28x^2-13\)

\(< =>4x^3-\left(4x^2+28x^2\right)+8x+13=0\)

\(< =>4x^3-32x^2+8x+13=0\)

do cái này nghiệm màu mè nên mình sẽ làm cách khá khó hiểu 

\(< =>x^3-7x^2+2x+\frac{13}{4}=0\)

Đặt \(x=y+\frac{7}{3}\)khi đó phương trình trở thành 

\(\left(y+\frac{7}{3}\right)^3-7\left(y+\frac{7}{3}\right)^2+2\left(y+\frac{7}{3}\right)+\frac{13}{4}=0\)

\(< =>y^3+3y^2\frac{7}{3}+3y\frac{49}{9}-7\left(y^2+\frac{14y}{3}+\frac{49}{9}\right)+2y+\frac{14}{3}+\frac{13}{4}=0\)

\(< =>y^3+7y^2+\frac{49y}{3}-7y^2-\frac{98y}{3}-\frac{343}{9}+2y+\frac{95}{12}=0\)

\(< =>y^3-\frac{43y}{3}-\frac{1087}{36}=0\)

Đặt \(y=u+v\)sao cho \(uv=\frac{43}{9}\)khi đó pt trở thành 

\(\left(u+v\right)^3-\frac{43\left(u+v\right)}{3}-\frac{1087}{36}=0\)

\(< =>u^3+v^3+3uv\left(u+v\right)-\left(u+v\right).\frac{43}{3}-\frac{1087}{36}=0\)

\(< =>u^3+v^3+\left(u+v\right)\left(3uv-\frac{43}{3}\right)-\frac{1087}{36}=0\)

\(< =>u^3+v^3=\frac{1087}{36}\)(*) (do \(uv=\frac{43}{9}\Leftrightarrow3uv-\frac{43}{3}=0\))

Ta có \(uv=\frac{43}{9}\Leftrightarrow u^3v^3=\frac{79507}{729}\)(**)

Từ (*) và (*) Suy ra được \(\hept{\begin{cases}u^3+v^3=\frac{1087}{36}\\u^3v^3=\frac{79507}{729}\end{cases}}\)

đến đây dễ rồi nhé ^^

a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)

\(=8x^5+2x^4-6x^3-14x^2\)

b: \(=2x^3-3x^2-5x+6x^2-9x-15\)

\(=2x^3+3x^2-14x-15\)

c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)

d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)

e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)

=2x^2-5x+1

a: ĐKXĐ: \(x\in R\)

\(\sqrt{\left(2x+3\right)^2}=5\)

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

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

b: ĐKXĐ: \(x\in R\)

\(\sqrt{9\left(x-2\right)^2}=18\)

=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)

=>\(3\cdot\left|x-2\right|=18\)

=>\(\left|x-2\right|=6\)

=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2

\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)

=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)

=>\(4\sqrt{x-2}=40\)

=>\(\sqrt{x-2}=10\)

=>x-2=100

=>x=102(nhận)

d: ĐKXĐ: \(x\in R\)

\(\sqrt{4\left(x-3\right)^2}=8\)

=>\(\sqrt{\left(2x-6\right)^2}=8\)

=>|2x-6|=8

=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)

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

e: ĐKXĐ: \(x\in R\)

\(\sqrt{4x^2+12x+9}=5\)

=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)

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

=>|2x+3|=5

=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)

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

f: ĐKXĐ:x>=6/5

\(\sqrt{5x-6}-3=0\)

=>\(\sqrt{5x-6}=3\)

=>\(5x-6=3^2=9\)

=>5x=6+9=15

=>x=15/5=3(nhận)

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

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

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

sai đề rồi bạn ơi