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C = -( 9x2 -2x +1) -17
= -(3x-1)2-17
ta có -(3x-1)2 bé hơn hoặc bằng 0 với mọi x
nên -(3x-1)2 -17 bé hơn hoặc bằng -17 với mọi x
vậy.............
\(C=-9x^2+2x-17\)
\(=-9\left(x^2-2.\dfrac{1}{9}x+\dfrac{1}{81}\right)-\dfrac{152}{9}\)
\(=-9\left(x-\dfrac{1}{9}\right)^2-\dfrac{152}{9}\)
Vì \(-9\left(x-\dfrac{1}{9}\right)^2\le0\)
Nên \(-9\left(x-\dfrac{1}{9}\right)^2-\dfrac{152}{2}\le0\)
Vậy C luôn âm với mọi giá trị của biến
\(D=-5x^2-6x-11\)
\(=-5\left(x^2+2.\dfrac{3}{5}x+\dfrac{9}{25}\right)-\dfrac{46}{5}\)
\(=-5\left(x+\dfrac{3}{5}\right)^2-\dfrac{46}{5}\)
Vì \(-5\left(x+\dfrac{3}{5}\right)^2\le0\)
Nên \(-5\left(x+\dfrac{3}{5}\right)^2-\dfrac{46}{5}\le0\)
vậy D luôn âm với mọi giá trị của biến
\(E=\dfrac{-1}{4}x^2+3x-15\)
\(=-\dfrac{1}{4}\left(x^2-12x+36\right)-6\)
\(=-\dfrac{1}{4}\left(x-6\right)^2-6\le0\)
Vậy E luôn âm với mọi giá trị
a) \(\frac{6-x}{3}-\frac{x}{4}=\frac{3+2x}{2}-1\)
\(\frac{4\left(6-x\right)}{12}-\frac{3x}{12}=\frac{3+2x}{2}-\frac{2}{2}\)
\(\frac{24-4x-3x}{12}=\frac{3+2x-2}{2}\)
\(\frac{24-7x}{12}=\frac{2x+1}{2}\)
\(\Rightarrow2\left(24-7x\right)=12\left(2x+1\right)\)
\(\Rightarrow48-14x=24x+12\)
\(\Rightarrow24x+14x=48-12\)
\(\Rightarrow38x=36\)
\(\Rightarrow x=\frac{18}{19}\)
b) \(-7x-\frac{x-3}{5}-\frac{x}{2}=x+\frac{2x+1}{3}\)
\(\frac{-70x}{10}-\frac{2\left(x-3\right)}{10}-\frac{5x}{10}=\frac{3x}{3}+\frac{2x+1}{3}\)
\(\frac{-70x-2x+6-5x}{10}=\frac{3x+2x+1}{3}\)
\(\frac{-77x+6}{10}=\frac{5x+1}{3}\)
\(\Rightarrow3\left(-77x+6\right)=10\left(5x+1\right)\)
\(\Leftrightarrow-231x+18=50x+10\)
\(\Leftrightarrow50x+231x=18-10\)
\(\Leftrightarrow281x=8\)
\(\Leftrightarrow x=\frac{8}{281}\)
Mấy câu kia tương tự
Điều kiện \(\hept{\begin{cases}x\ne0\\3x^2-x-4\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{4}{3}\end{cases}}}\)
Đặt \(\frac{3x^2-x-4}{x}=a\)thì ta có
\(PT\Leftrightarrow a+\frac{9}{a}=6\)
\(\Leftrightarrow a^2-6a+9=0\)
\(\Leftrightarrow\left(a-3\right)^2=0\)
\(\Leftrightarrow a=3\)
\(\Leftrightarrow\frac{3x^2-x-4}{x}=3\)
\(\Leftrightarrow3x^2-4x-4=0\)
\(\Leftrightarrow\left(3x^2-6x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}\)
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
4)a)\(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)(1)
ĐKXĐ:\(\left\{{}\begin{matrix}x-5\ne0\\x+5\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-5\end{matrix}\right.\)
(1)\(\Rightarrow\left(x+5\right)\left(x+5\right)-\left(x-5\right)\left(x-5\right)=20\)
\(\Leftrightarrow x^2+10x+25-\left(x^2-10x+25\right)=20\)
\(\Leftrightarrow x^2+10x+25-x^2+10x-25=20\)
\(\Leftrightarrow x^2-x^2+10x+10x=-25+25=20\)
\(\Leftrightarrow20x=20\)
\(\Leftrightarrow x=1\left(nh\text{ậ}n\right)\)
S=\(\left\{1\right\}\)
`5+76/[x^2-16]=[2x-1]/[x+4]-[3x-1]/[4-x]` `ĐK: x \ne +-4`
`<=>[5(x^2-16)+76]/[(x-4)(x+4)]=[(2x-1)(x-4)+(3x-1)(x+4)]/[(x+4)(x-4)]`
`=>5x^2-80+76=2x^2-8x-x+4+3x^2+12x-x-4`
`<=>2x=-4`
`<=>x=-2` (t/m)
Vậy `S={-2}`