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Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết
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1, \(\left(x-4\right)^2-\left(2x+1\right)^2=\left(x-4-2x-1\right)\left(x-4+2x+1\right)=-3\left(x+5\right)\left(x-1\right).\)
\(\orbr{\begin{cases}x+5=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=1\end{cases}}}\)(mấy cái này áp dụng hàng đẳng thức lớp 8 mới hok)
2,\(x^3+x^2-4x-4=\left(x-2\right)\left(x^2+3x+2\right)=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(\orbr{\begin{cases}x=\mp2\\\end{cases}}x=-1\)
tương tụ lm tiếp nhe buồn ngủ quá rồi !
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4) \(2.3^x+3^{x-1}=7.\left(3^2+2.6^2\right)\)
\(\Rightarrow2.3^x+3^{x-1}=567\)
\(\Rightarrow7.3^{x-1}=567\)
\(\Rightarrow3^{x-1}=567\div7\)
\(\Rightarrow3^{x-1}=81\)
\(\Rightarrow3^{x-1}=3^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=4+1\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
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4 x 3x-1 + 2 x 3x+2 = 4 x 36 +2 x 39
=> 3x-1 = 36 => x - 1 = 6 => x = 6 + 1 = 7
=> 3x+2 = 39 => x + 2 = 9 => x = 9 - 2 = 7
Vậy x = 7
\(4.3^{x-1}+2.3^{x+2}=4.3^6+2.3^9\)
\(\Rightarrow2.3^{x-1}\left(2+27\right)=2.3^6\left(2+27\right)\)
\(\Rightarrow2.3^{x-1}=2.3^6\)
\(\Rightarrow x-1=6\Leftrightarrow x=7\)
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a) \(x+\frac{1}{2}=2^5:2^3\)
\(x+\frac{1}{2}=4\)
\(x=4-\frac{1}{2}\)
\(x=\frac{7}{2}\)
b)\(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\)
\(\frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\)
\(\frac{5}{3}x=\frac{1}{21}\)
\(x=\frac{1}{21}:\frac{5}{3}\)
\(x=\frac{1}{35}\)
c)\(\left|x+5\right|-6=9\)
\(\left|x+5\right|=9+6\)
\(\left|x+5\right|=15\)
\(\Rightarrow\orbr{\begin{cases}x+5=15\\x+5=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=15-5\\x=\left(-15\right)-5\end{cases}}\Rightarrow\orbr{\begin{cases}x=10\\x=-20\end{cases}}}\)
Vậy x = 10 ; x= -20
d)\(-\frac{12}{13}x-5=6\frac{1}{3}\)
\(-\frac{12}{13}x-5=\frac{19}{3}\)
\(-\frac{12}{13}x=\frac{19}{3}+5\)
\(-\frac{12}{13}x=\frac{34}{3}\)
\(x=\frac{34}{3}:\left(-\frac{12}{13}\right)\)
\(x=-\frac{221}{18}\)
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c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
\((x-\dfrac{2}{9})^3=(\dfrac{2}{3})^6\)
\(=>(x-\dfrac{2}{9})^3=[(\dfrac{2}{3})^2]^3\)
\(=>(x-\dfrac{2}{9})^3=(\dfrac{4}{9})^3\)
\(=>x-\dfrac{2}{9}=\dfrac{4}{9}\)
\(=>x=\dfrac{4}{9}+\dfrac{2}{9}=\dfrac{2}{3}\)
(x - 2/9)3 = (2/3)6
(x - 2/9)3 = [(2/32)]3
(x - 2/9)3 = (4/9)3
x - 2/9 = 4/9
x = 4/9 + 2/9
x = 6/9
x = 2/3
Vậy x = 2/3