2x/7=y/3;x/2=2z/5 và x+y-z= 34
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\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
a: =>-0,5x+1,5=0,4x-0,2
=>-0,9x=-1,7
=>x=17/9
3x-1/2x+3=3x+2/2x-1
=>6x^2-3x-2x+1=6x^2+4x+9x+6
=>-5x+1=13x+6
=>-8x=5
=>x=-5/8
b: \(\Leftrightarrow\left(4x-1\right)\left(-x+7\right)=\left(4x+5\right)\left(-x-2\right)\)
=>\(-4x^2+28x+x-7=-4x^2-8x-5x-10\)
=>29x-7=-13x-10
=>42x=-3
=>x=-1/14
c: =>7x=5y và 2x-y=15
=>7x-5y=0 và 2x-y=15
=>x=25; y=35
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\y\ge1\end{matrix}\right.\)
\(xy-y^2+2y-x-1=\sqrt{y-1}-\sqrt{x}\)
\(\Leftrightarrow x\left(y-1\right)-\left(y-1\right)^2+\sqrt{x}-\sqrt{y-1}=0\)
\(\Leftrightarrow\left(y-1\right)\left(x-y+1\right)+\dfrac{x-y+1}{\sqrt{x}+\sqrt{y-1}}=0\)
\(\Leftrightarrow\left(x-y+1\right)\left(y-1+\dfrac{1}{\sqrt{x}+\sqrt{y-1}}\right)=0\)
\(\Leftrightarrow x-y+1=0\)
\(\Rightarrow y=x+1\)
Thay xuống pt dưới:
\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)
\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+\left(7-x-3\sqrt{5-x}\right)=0\)
\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)
\(\Leftrightarrow...\)
đk x khác 7/2 ; y khác -6
Đặt \(\dfrac{1}{2x-7}=t;\dfrac{1}{y+6}=u\)
\(\left\{{}\begin{matrix}3t+4u=7\\2t-3u=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6t+8u=14\\6t-9u=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=1\\t=\dfrac{-1+3u}{2}=1\end{matrix}\right.\)
Theo cách đặt \(\left\{{}\begin{matrix}2x-7=1\\y+6=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-5\end{matrix}\right.\left(tm\right)\)
làm giúp mk bài này nhá 0+1+2+...+2017 có bao nhiêu số hạng
Giải hệ phương trình: \(\left\{\begin{matrix} xy-y^2-x+2y=\sqrt{y-1}+1-\sqrt{x} - Hy Vũ