b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)

\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)

\(\Leftrightarrow x^2+7x+6=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

\(a,2x\left(x+5\right)=\left(x+3\right)^2+\left(x-1\right)^2+20\)
\(\Leftrightarrow2x^2+10x=x^2+6x+9+x^2-2x+1+20\)
\(\Leftrightarrow2x^2-x^2-x^2+10x-6x+2x=30\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)

\(b,\left(2x-2\right)^2=\left(x+1\right)^2+3\left(x-2\right)\left(x+5\right)\)

\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+3x-10\right)\)

\(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3x^2+9x-30\)

\(\Leftrightarrow4x^2-8x-x^2-3x^2-2x-9x=-33\)

\(\Leftrightarrow-19x=-33\)

\(\Leftrightarrow x=\frac{33}{19}\)

\(c,\left(x-1\right)^2+\left(x+3\right)^2=2\left(x-2\right)\left(x+1\right)+38\)

\(\Leftrightarrow x^2-2x+1+x^2+6x+9=2\left(x^2-x-2\right)+38\)

\(\Leftrightarrow6x=25\)

\(\Leftrightarrow x=\frac{25}{6}\)

a) \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x=-x\left(x+3\right)=-3\left(3+3\right)=-18\)

b) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2=3\left(x^2-\frac{4}{5}y^2\right)\)

\(=3\left(4^2-\frac{4}{5}.5^2\right)=3.\left(-4\right)=-12\)

c) \(\left(x-2\right)^2-\left(x+7\right)\left(x-7\right)=x^2-4x+4-x^2+49=-4x+53=-4.3+53=41\)

d) \(x^2+12x+36=\left(x+6\right)^2=\left(64+6\right)^2=70^2=4900\)

e) \(\left(x-3\right)^2-\left(x-4\right)\left(x+4\right)=x^2-6x+9-x^2+16=-6x+25=-6\left(-1\right)+25\)

= 31

f) \(\left(3x+2y\right)^2-4y\left(3x+y\right)=9x^2+12xy+4y^2-12xy-4y^2=9x^2=9\left(-\frac{1}{3}\right)^2=1\)

a, \(5x^2-2x\left(3x+\frac{3}{2}\right)=-x^2-3x\)

Thay x = 3 vào biểu thức trên ta cs : \(-3^2-3.3=-9-9=-18\)

b, \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-\frac{12}{5}y^2\)

Thay x = 4 ; y = 5 vào biểu thức trên ta có : \(3.4^2-\frac{12}{5}.5^2=-12\)

4, \(\Leftrightarrow4x+4+9\left(2x+1\right)=4x+6\left(x+1\right)+7+12x\)

\(\Leftrightarrow22x+13=22x+13\)vậy pt có vô số nghiệm 

5, \(\dfrac{2x}{3}+\dfrac{2x-1}{6}=4-\dfrac{x}{3}\Rightarrow4x+2x-1=24-2x\)

\(\Leftrightarrow8x=25\Leftrightarrow x=\dfrac{25}{8}\)

6, \(\dfrac{x-1}{2}+\dfrac{x-1}{4}=1-\dfrac{2\left(x-1\right)}{3}\Rightarrow6x-6+3x-3=12-8\left(x-1\right)\)

\(\Leftrightarrow9x-9=20-8x\Leftrightarrow17x=29\Leftrightarrow x=\dfrac{29}{17}\)

vào câu hỏi tương tự mà lm tương tự như thế nha

a/ ĐKXĐ: ...

\(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}-\frac{5x-9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{1}{6x}\left(1-\frac{5x-9}{5x-1}\right)=\frac{8x-5}{8x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{4}{3x\left(5x-1\right)}-\frac{8x-5}{8x\left(5x+1\right)}=0\)

\(\Leftrightarrow24x\left(5x-3\right)+32\left(5x+1\right)-3\left(5x-1\right)\left(8x-5\right)=0\)

\(\Leftrightarrow-120x^2+379x-55=0\)

Bạn có nhầm đề chỗ nào ko nhỉ? Con số thật khủng khiếp (nghiệm ko hề đẹp)

b/ ĐKXĐ:...

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\right)=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=27-\frac{1}{x+9}\)

\(\Leftrightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)

a, \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\) (ĐKXĐ: x \(\ne\) \(\pm\)\(\frac{1}{5}\); x \(\ne\) 0)

\(\Leftrightarrow\) \(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}+\frac{-5x+9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\) \(\frac{24x\left(5x-3\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}+\frac{-4\left(5x+1\right)\left(5x-9\right)}{48x\left(5-1x\right)\left(5x+1\right)}+\frac{4\left(5x-1\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}=\frac{3\left(8x-5\right)\left(5x-1\right)}{48x\left(5x-1\right)\left(5x+1\right)}\)

\(\Leftrightarrow\) 24x(5x - 3) - 4(5x + 1)(5x - 9) + 4(5x - 1)(5x + 1) = 3(8x - 5)(5x - 1)

\(\Leftrightarrow\) 120x² - 72x - 100x² + 160x + 36 + 100x² - 4 = 120x² - 99x + 15

\(\Leftrightarrow\) 120x² - 120x² - 100x² + 100x² - 72x + 160x + 99x = 15 - 36 + 4

\(\Leftrightarrow\) 187x = -17

\(\Leftrightarrow\) x = \(\frac{-1}{11}\) (TM ĐKXĐ)

Vậy S = {\(\frac{-1}{11}\)}

Chúc bn học tốt!! (Đã được kiểm chứng không sai :)

b, \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -3; x \(\ne\) -6; x \(\ne\) -9)

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) - \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}-27+\frac{1}{x+9}\)) = 0

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-27\)) = 0

\(\Leftrightarrow\) \(\frac{1}{x}-27\) = 0

\(\Leftrightarrow\) x = \(\frac{1}{27}\) (TM ĐKXĐ)

Vậy S = {\(\frac{1}{27}\)}

Chúc bn học tốt!!

1: A=4x^2+12x+9-4x^2+4x-1-6x=10x+8

Khi x=201 thì A=10*201+8=2018

2: B=4x^2+20x+25-4x^2+12=20x+37

Khi x=1/20 thì B=1+37=38

1, \(A=\left(2x+3\right)^2-\left(2x-1\right)^2-6x\)

\(A=\left[\left(2x+3\right)+\left(2x-1\right)\right]\left[\left(2x+3\right)-\left(2x-1\right)\right]-6x\)

\(A=\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)-6x\)

\(A=4\left(4x+2\right)-6x\)

\(A=16x+8-6x\)

\(A=10x+8\)

Thay \(x=201\) vào A ta có:

\(A=10\cdot201+8=2010+8=2018\)

Vậy: ....

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

\(B=\left(2x+5\right)^2-4\left(x^2-9\right)\)

\(B=4x^2+20x+25-4x^2+36\)

\(B=20x+61\)

Thay \(x=\dfrac{1}{20}\) vào B ta có:

\(B=20\cdot\dfrac{1}{20}+61=1+61=62\)

Vậy: ...