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1)
\(\Leftrightarrow\left(x^2-2+\dfrac{1}{x^2}\right)+\left(y^2-2+\dfrac{1}{y^2}\right)+z^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2+z^2=0\)
\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\Rightarrow\left|x\right|=1\\y-\dfrac{1}{y}=0\Rightarrow\left|y\right|=1\\z=0\end{matrix}\right.\)
dk\(x,y,z,a,b,c\ne0\)\(\left\{{}\begin{matrix}\dfrac{a}{x}=A\\\dfrac{b}{y}=B\\\dfrac{c}{z}=C\end{matrix}\right.\) \(\Rightarrow A,B,C\ne0\)
\(\left\{{}\begin{matrix}A+B+C=2\\\dfrac{1}{A}+\dfrac{1}{B}+\dfrac{1}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}A^2+B^2+C^2+2\left(AB+BC+AC\right)=4\\\dfrac{ABC}{A}+\dfrac{ABC}{B}+\dfrac{ABC}{C}=0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}AB+BC+AC=0\\A^2+B^2+C^2=4\end{matrix}\right.\)
\(\left(\dfrac{a}{x}\right)^2+\left(\dfrac{b}{y}\right)^2+\left(\dfrac{c}{z}\right)^2=4\)
\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)
\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)
\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
Bạn áp dụng 7 hằng đẳng thức ta đã học từ đầu năm học lớp 8 là ra nhé
a )
\(\left(1+3a\right)^2=9a^2+6a+1\)
b )
\(\left(2a+3\right)\left(2a-3\right)=4a^2-9\)
c )
\(\left(2a^2+b^2\right)^2=4a^4+4a^2b^2+b^4\)
d )
\(\left(\dfrac{a}{2}-2b\right)^2=\dfrac{a^2}{4}-2ab+4b^2\)
e )
\(\left(a^2+5\right)\left(5-a^2\right)=25-a^2\)
f )
\(\left(\dfrac{1}{2}a-2b\right)^3=\dfrac{1}{8}a^3-\dfrac{3}{2}a^2b+6ab^2-8b^3\)
Chúc bạn học tốt !!
\(P=\dfrac{\left(2a-1\right)\left(3a+1\right)+\left(5-a\right)\left(3a-1\right)}{\left(3a-1\right)\left(3a+1\right)}\)
\(=\dfrac{6a^2-a-1-3a^2+16a-5}{9a^2-1}=\dfrac{3a^2+15a-6}{9a^2-1}\)
\(=\dfrac{\left(30a^2+15a\right)-\left(27a^2-3\right)-9}{9a^2-1}\)
\(=\dfrac{3\left(10a^2+5a\right)-9-3\left(9a^2-1\right)}{9a^2-1}\)
\(=\dfrac{3.3-9-3\left(9a^2-1\right)}{9a^2-1}=-3\) (vì 10a2 + 5a = 3)