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Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{c}{z}=m\)ta có:
\(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{m}+\frac{1}{m}=2\)
\(\Leftrightarrow\frac{3}{m}=2\)
\(\Leftrightarrow m=1,5\)
bạn à!
đề bài là giải phương trình trên nhá lúc đánh mình quên mất
c, Ta có : \(2x^2+2x+3x+3=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy ...
d, Ta có : \(\dfrac{3-2x}{2006}+\dfrac{3-2x}{2007}+\dfrac{3-2x}{2008}=\dfrac{3-2x}{2009}+\dfrac{3-2x}{2010}\)
\(\Leftrightarrow\dfrac{3-2x}{2006}+\dfrac{3-2x}{2007}+\dfrac{3-2x}{2008}-\dfrac{3-2x}{2009}-\dfrac{3-2x}{2010}=0\)
\(\Leftrightarrow\left(3-2x\right)\left(\dfrac{1}{2006}+\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}-\dfrac{1}{2010}\right)=0\)
\(\Leftrightarrow3-2x=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy ...
a) Ta có: \(\left(3x-2\right)\left(4x+3\right)=\left(2-3x\right)\left(x-1\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(4x+3\right)-\left(2-3x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x+3\right)+\left(3x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x+3+x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\5x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\5x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{2}{5}\right\}\)
b) Ta có: \(x^2+\left(x+3\right)\left(5x-7\right)=9\)
\(\Leftrightarrow x^2+5x^2-7x+15x-21-9=0\)
\(\Leftrightarrow6x^2+8x-30=0\)
\(\Leftrightarrow6x^2+18x-10x-30=0\)
\(\Leftrightarrow6x\left(x+3\right)-10\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(6x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-3;\dfrac{5}{3}\right\}\)
b) Ta có : a\(^2\)+ b\(^2\)+ c\(^2\) =ab+bc+ca
=> 2(a\(^2\)+b\(^2\)+c\(^2\))= 2(ab+bc+ca)
<=>2a\(^2\)+2b\(^2\)+2c\(^2\)=2ab+2bc+2ca
<=> 2a\(^2\)+2b\(^2\)+2c\(^2\)-2ab-2bc-2ca=0
<=> a\(^2\)+a\(^2\)+b\(^2\)+b\(^2\)+c\(^2\)+c\(^2\)-2ab-2bc=2ca=0
<=> (a\(^2\)-2ab+b\(^2\))+(b\(^2\)-2bc+b\(^2\))+(a\(^2\)-2ca+c\(^2\))
<=> (a-b)\(^2\)+(b-c)\(^2\)+(a-c)\(^2\) =a
<=> hoặc a-b=0 hoặc b-c=o hoặc a-c=o <=>a=b hoặc b=c hoặc a=c
=>a=b=c (đpcm)
a) Theo đề bài: \(a^2+b^2=ab\)
=>\(a^2+b^2-ab=0\)
=>\(a^2-2ab+b^2+ab=0\)
=>\(\left(a-b\right)^2+ab=0\)
Vì \(\left(a-b\right)^2\ge0\) để \(\left(a-b\right)^2+ab=0\) <=> \(\left(a-b\right)^2=ab=0\)
(a-b)2=0 <=> a-b=0 <=> a=b (đpcm)
b)\(a^2+b^2+c^2=ab+bc+ca\)
=>\(2\left(a^2+b^2+c^2\right)=2\left(ab+bc+ac\right)\)
=>\(2a^2+2b^2+2c^2=2ab+2bc+2ac\)
=>\(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
=>\(\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)=0\)
=>\(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
Vì \(\begin{cases}\left(a-b\right)^2\ge0\\\left(b-c\right)^2\ge0\\\left(a-c\right)^2\ge0\end{cases}\) để \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
<=>\(\left(a-b\right)^2=\left(b-c\right)^2=\left(a-c\right)^2=0\)
<=>a-b=b-c=a-c=0
<=>a=b=c (đpcm)
bai kho wa zay ban oi
mình k biết làm nhưng bạn thử gõ lên google thử xem ! biết đâu sẽ có đấy :)