Đặt \(\hept{\begin{cases}\sqrt{7x+11}=a\\\sqrt{9-7x}=b\end{cases}}\)

\(\Rightarrow a^2-b^2=14x+2\)

\(\Rightarrow\frac{2}{a^2-b^2}+\frac{1}{ab}=\frac{7}{24}\)

\(\Leftrightarrow\left(b+7a\right)\left(7b-a\right)=0\)

Làm nhầm phần phân tích nhân tử giờ làm lại cách khác.

Đặt \(7x+11=a\)

\(\Rightarrow7x=a-11\)

\(\Rightarrow\frac{1}{a-10}+\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}\)

\(\Leftrightarrow\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}-\frac{1}{a-10}\)

\(\Leftrightarrow\frac{1}{a\left(20-a\right)}=\left(\frac{7}{24}-\frac{1}{a-10}\right)^2\)

\(\Leftrightarrow\left(a-18\right)\left(a-16\right)\left(49a^2-630a+200\right)=0\)

PS: Bài giải trên bỏ đi nha

\(N=\dfrac{\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)+1}{x^2+7x+11}\)

\(=\dfrac{\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]+1}{x^2+7x+11}\)

\(=\dfrac{\left(x^2+7x+10\right)\left(x^2+7x+12\right)+1}{x^2+7x+11}\)

Đặt \(x^2+7x+11=y\), thay vào \(N\) ta được:

\(N=\dfrac{\left(y-1\right)\left(y+1\right)+1}{y}\)

\(=\dfrac{y^2-1+1}{y}\)

\(=\dfrac{y^2}{y}\)

\(=y\)

\(=x^2+7x+11\)

Vậy \(N=x^2+7x+11\).

\(\text{#}Toru\)

A=(x+1)(x+2)(x+3)(x+4)-24

=(x²+5x+4)(x²+5x+6)-24

Đặt t=(x²+5x+4) ta có:

t(t+2)-24=t²+6t-2t-24

=t(t+6)-4(t+6)

=(t-4)(t+6).Thay vào ta đc:

(x²+5x+4-4)(x²+5x+4+6)=(x²+5x)(x²+5x+10) 

=x(x+5)(x²+5x+10)

B=(x²+3x+2)(x²+7x+120-24)

=(x²+3x+2)(x²+7x+96)

=(x²+2x+x+2)(x²+7x+96)

=[x(x+2)+(x+2)](x²+7x+96)

=(x+1)(x+2)(x²+7x+96)

C và D bn cx lm tương tự

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Leftrightarrow\left(7x-11\right)^3=1000\)

\(\Leftrightarrow7x-11=10\)

\(\Leftrightarrow7x=21\)

\(\Leftrightarrow x=3\)

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=800+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

\(\left(7x-11\right)^3=2^5.5^2+200\)

\(\Rightarrow\left(7x-11\right)^3=32.25+200\)

\(\Rightarrow\left(7x-11\right)^3=800+200\)

\(\Rightarrow\left(7x-11\right)^3=1000=10^3\)

\(\Rightarrow7x-11=10\)

\(\Rightarrow7x=21\)

\(\Rightarrow x=3\)

( 7x - 11 ) ³ = ( -3 )² .15 + 208

( 7x - 11 ) ³ = 9 .15 + 208

( 7x - 11 ) ³ = 135 + 208

( 7x - 11 ) ³ = 343

( 7x - 11 ) ³ = 7³

=> 7x - 11 = 7

=> 7x = 7 + 11 = 18

=> x = 18/7

\(\left(7x-11\right)^3=\left(-3\right)^2.15+208\)

\(\Leftrightarrow\)\(\left(7x-11\right)^3=9.15+208\)

\(\Leftrightarrow\)\(\left(7x-11\right)^3=135+208\)

\(\Leftrightarrow\)\(\left(7x-11\right)^3=343\)

\(\Leftrightarrow\)\(\left(7x-11\right)^3=7^3\)

\(\Leftrightarrow\)\(7x-11=7\)

\(\Leftrightarrow\)\(7x=18\)

\(\Leftrightarrow\)\(x=\frac{18}{7}\)

Vậy \(x=\frac{18}{7}\)

Chúc bạn học tốt ~ 

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

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=8\left(7x+4\right)\)

=56x+32

b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)

\(=8x^2-32x+32-3x^2+12x+15-5x^2\)

\(=-20x+47\)

c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)

\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)

=2

\(x\inℤ\) hả 

x thuộc Z

CHỜ TÍ

ko biết

1.

\(x-4=11-2x\)

\(\Leftrightarrow x+2x=11+4\)

\(\Leftrightarrow 3x=15\Leftrightarrow x=15:3=5\)

2.

\((x-3)(x-2)=0\Rightarrow \left[\begin{matrix} x-3=0\\ x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=2\end{matrix}\right.\)

3.

\(7x-10=x-28\)

\(\Leftrightarrow 7x-10-(x-28)=0\)

\(\Leftrightarrow 6x+18=0\Leftrightarrow 6x=-18\Leftrightarrow x=-3\)

4.

\(|x-5|=10-|-2|=10-2=8\)

\(\Rightarrow \left[\begin{matrix} x-5=8\\ x-5=-8\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=13\\ x=-3\end{matrix}\right.\)

a)\(\left\{{}\begin{matrix}8x+2y=4\\8x+3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\4x+1=2\end{matrix}\right.\Leftrightarrow}\left\{{}\begin{matrix}y=1\\x=\frac{1}{4}\end{matrix}\right.\)b)

\(\left\{{}\begin{matrix}12x-8y=44\\12x-15y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=35\\4x-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\4x-5.5=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=7\end{matrix}\right.\)c)\(\left\{{}\begin{matrix}9x=-18\\4x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\4.\left(-2\right)+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=7\end{matrix}\right.\)

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