(x-9)^4+(x-10)^4=(19-2x)^4
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Đặt \(\left\{{}\begin{matrix}x-9=a\\x-10=b\end{matrix}\right.\)
\(a^4+b^4=\left(a+b\right)^4\)
\(\Leftrightarrow a^4+b^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
\(\Leftrightarrow2ab\left(2a^2+3ab+2b^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\b=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=10\end{matrix}\right.\)
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`Answer:`
a. \(x^3+6x^2+12=19\)
\(\Leftrightarrow x^3+6x^2+12x-19=0\)
\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)
\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)
Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)
\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)
\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)
\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)
\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)
\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)
\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)
c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)
\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(2x+3-x+2\right)^2\)
\(=\left(x+5\right)^2\)
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a. y4 - 14y2 + 49
Gọi y2 là t, ta có:
t2 - 14t + 49
<=> t2 - 14t + 72
<=> (t - 7)2
Thay x2 = t
<=> (x2 - 7)2
b. \(\dfrac{1}{4}-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^2-x^2\)
\(\Leftrightarrow\left(\dfrac{1}{2}-x\right)\left(\dfrac{1}{2}+x\right)\)
c. x4 - 16
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
d. x2 - 9
<=> x2 - 32
<=> (x - 3)(x + 3)
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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
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a) Ta có: \(3-\left(17-x\right)=-12\)
\(\Leftrightarrow3-17+x+12=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: x=2
b) Ta có: \(\left(2x+4\right)\left(10-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=0\\10-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-4\\2x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;5\right\}\)c) Ta có: \(\left|x-9\right|=-2+17\)
\(\Leftrightarrow\left|x-9\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=15\\x-9=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=24\\x=-6\end{matrix}\right.\)
Vậy: \(x\in\left\{24;-6\right\}\)