a, \(\left(x-\dfrac{3}{4}\right)^2=0\Rightarrow x=\dfrac{3}{4}\)

Vậy...

b, \(\left(x-3\right)^2=1\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

Vậy x = 4 hoặc x = 2

c, \(\left(2x+1\right)^3=-8\)

\(\Rightarrow2x+1=-3\)

\(\Rightarrow x=-2\)

Vậy x = -2

d, \(\left(x-\dfrac{1}{4}\right)^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{4}=\dfrac{1}{2}\\x-\dfrac{1}{4}=\dfrac{-1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{-1}{4}\end{matrix}\right.\)

Vậy...

1, \(\left(x-6\right)^2-x^2+36\)

\(=\left(x-6\right)^2-\left(x-6\right)\left(x+6\right)\)

\(=\left(x-6\right)\left(x-6-x-6\right)\)

\(=-12\left(x-6\right)\)

2, \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-yz-zx+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)

3, sai đề

\(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{5}{24}\\\dfrac{9}{x}+\dfrac{6}{5}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)=1\end{matrix}\right.\)

\(\Rightarrow\dfrac{9}{x}+\dfrac{6}{5}.\dfrac{5}{24}=1\)

\(\Leftrightarrow\dfrac{9}{x}+\dfrac{1}{4}=1\)

\(\Leftrightarrow\dfrac{9}{x}=\dfrac{3}{4}\)

\(\Leftrightarrow3x=36\Leftrightarrow x=12\)

\(\Rightarrow\dfrac{1}{12}+\dfrac{1}{y}=\dfrac{5}{24}\)

\(\Rightarrow\dfrac{1}{y}=\dfrac{1}{8}\Rightarrow y=8\)

Vậy x = 12, y = 8

Ta có:

\(\dfrac{3}{2}x=\dfrac{4}{3}y=\dfrac{5}{4}z\Rightarrow\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{2}{3}}=\dfrac{y}{\dfrac{3}{4}}=\dfrac{z}{\dfrac{4}{5}}=\dfrac{2y}{\dfrac{3}{2}}=\dfrac{x-2y+z}{\dfrac{2}{3}-\dfrac{3}{2}+\dfrac{4}{5}}=\dfrac{-16}{\dfrac{-1}{30}}=480\)

\(\Rightarrow\left\{{}\begin{matrix}x=320\\y=360\\z=384\end{matrix}\right.\)

Vậy...

Bài 2:

\(\left(x+1\right)^4=\left(x+1\right)^2\)

\(\Rightarrow\left(x+1\right)^4-\left(x+1\right)^2=0\)

\(\Rightarrow\left(x+1\right)^2\left[\left(x+1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\\left(x+1\right)^2-1=0\end{matrix}\right.\)

+) \(\left(x+1\right)^2=0\Rightarrow x=-1\)

+) \(\left(x+1\right)^2-1=0\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{-1;0;-2\right\}\)

\(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy...

a, Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow\left\{{}\begin{matrix}a=3k\\b=4k\\c=5k\end{matrix}\right.\)

Ta có: \(4\left(a-b\right)\left(b-c\right)\)

\(=4\left(3k-4k\right)\left(4k-5k\right)\)

\(=4.\left(-k\right).\left(-k\right)=4k^2\) (1)

\(\left(a-c\right)^2=\left(3k-5k\right)^2=4k^2\) (2)

Từ (1), (2) \(\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)

\(\Rightarrowđpcm\)

tick cho minh với minh mới dk 0 điểm

a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)\)

\(=2000^2-1< 2000^2\)

\(\Rightarrow A< B\)

b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1< 2^{32}\)

\(\Rightarrow A< B\)