a ) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=24\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=24\)

\(\Leftrightarrow2x=-231\Leftrightarrow x=\dfrac{-231}{2}\)

b ) \(\left(x+3\right)^2-\left(x-4\right)\left(x-8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2+12x-32=1\)

\(\Leftrightarrow18x=24\Leftrightarrow x=\dfrac{4}{3}\)

Chúc bạn học tốt !!!!!!!!!!!!

a)\(10\left(x-7\right)-8\left(x+5\right)=6\cdot\left(-5\right)+24\)

\(10x-10\cdot7-8x-8\cdot5=\left(-30\right)+24\)

\(10x-70-8x-40=-6\)

\(10x-8x=\left(-6\right)+70+40\)

\(2x=104\)

\(x=104\div2\)

\(x=52\)

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

\(2\cdot4x-2\cdot8-7\cdot3-7x=6\)

\(8x-16-21-7x=6\)

\(8x-7x=6+16+21\)

\(x=43\)

câu a tự quy đồng cùng  mẫu rồi làm thôi :"))

b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)

\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)

Đặt \(x^2-x=k\), ta có:

\(k.\left(k-2\right)=24\)

\(\Leftrightarrow k^2-2k+1=25\)

\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)

\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)

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

\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)

c)\(x^4+2x^3+5x^2+4x-12=0\)

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

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

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

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

\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)

p/s: bn tự kết luận nha :))

\(\Leftrightarrow\)\(x-\left(\frac{13x}{18}-\frac{4}{18}\right)=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{18x}{18}-\frac{13x}{18}+\frac{4}{18}=\frac{4}{9}\)

\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{4}{9}-\frac{4}{18}\)

\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{2}{9}\)

\(\Leftrightarrow\)\(5x=\frac{18.2}{9}\)

\(\Leftrightarrow\)\(5x=4\)

\(\Leftrightarrow\)\(x=\frac{4}{5}\)

x³ + 3x² + 3x = 7

<=> x³ + 3x² + 3x - 7 = 0

<=> (x - 1)(x² + 4x + 7) = 0

<=> x - 1 = 0 hoặc x² + 4x + 7 khác 0

<=> x - 1 = 0

<=> x = 1

a) ( x² + 3 x + 2 ) . ( x² + 3x+ 3 ) - 2 =0

<=>x⁴ + 3x³ + 3x² + 3x³ + 9x² + 9x + 2x² + 6x + 6 - 2 = 0

<=> x⁴ +  6x³  + 14x² + 15x + 4 = 0

<=> x⁴ + 3x³ + 3x³ + x² + 9x² + 4x² + 3x + 12x + 4 = 0

<=> x² . ( x² +3x + 1 ) + 3x . ( x² +3x + 1 ) + 4. ( x² + 3x + 1 ) = 0

<=> ( x² + 3x + 1 ) . ( x² + 3x + 4 ) = 0

<=> \(\orbr{\begin{cases}x^2+3x+1=0\\x^2+3x+4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

          \(x\notinℝ\)

<=> \(\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=\frac{-3-\sqrt{5}}{2}\end{cases}}\)

Nghiệm cuối cùng là : x₁ = \(\frac{-3+\sqrt{5}}{2}\);x₂ = \(\frac{-3-\sqrt{5}}{2}\)

b) ( x + 1 ) . ( x + 2 ) . ( x + 3 ) . ( x + 4 )  - 24 = 0

<=> ( x² + 2x + x + 2 ) . ( x + 3 ) . ( x + 4 ) - 24 = 0

<=> ( x² + 3.x + 2 ) . ( x+3) . ( x + 4 ) -24         = 0

<=> ( x³ + 3.x ² + 3.x² + 9x + 2x + 6 ) . ( x + 4 ) - 24 = 0 

<=> ( x³ + 3x + 2 ) . ( x + 3 ) .( x + 4 ) = 0

<=> ( x³ + 3x² + 3x² + 9x + 2x + 6 ) . ( x + 4) - 24 = 0

<=> ( x³ + 6.x² + 11.x + 6 ) . ( x + 4 ) -24 = 0 

<=> x⁴ + 4.x³ + 6.x³ + 24.x² + 11.x² + 44.x + 6.x + 24 - 24 =0

<=> x⁴ + 10.x³+ 35. x²  + 50.x = 0

<=> x. ( x³ + 10.x² + 35 .x + 50 ) = 0

<=> x. ( x³ + 5.x² +5.x² + 25.x+ 10 + 50 ) = 0

<=> x. [ x² . ( x+5 ) + 5.x. ( x+5 ) + 10.( x + 5 ) ] = 0

<=> x. ( x + 5 ) . ( x² + 5.x + 10 ) = 0

=> \(\hept{\begin{cases}x=0\\x+5=0\\x^2+5.x+10=0\end{cases}}\)

=> \(\hept{\begin{cases}x=0\\x=-5\\x\notinℝ\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-5\\x=0\end{cases}}\)

Nghiệm cuối cùng là : x₁ = -5 ; x₂ = 0

c) x³ + 3.x² + 3x = 7

<=> x³ + 3.x² + 3x - 7 = 0

<=> ( x + 1 )³ - 8          = 0

<=> ( x + 1 )³               = 8

<=> ( x + 1 ) ³               = 2³ 

<=> x + 1                      = 2

<=> x                              =1

Vậy x = 1

ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A

a) \(\left(x-3\right)^2-4=0\)

\(\Rightarrow\left(x-3\right)^2=4\)

\(\Rightarrow\left(x-3\right)^2=2^2=\left(-2\right)^2\)

\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)

\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)

Vậy \(x\in\left\{5;-1\right\}\)

b) \(x^2-2x=24\)

\(\Rightarrow x.\left(x+2\right)=24\)

\(\Rightarrow x.\left(x+2\right)=4.6\)

\(\Rightarrow x=4\)

Vậy \(x=4\)