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d) x2 + 2x + 2 < 0
<=> x2 + 2x + 1 + 1 < 0
<=> ( x + 1 )2 + 1 < 0
<=> ( x + 1 )2 < -1 ( vô lí )
=> BPT vô nghiệm ( đpcm )
e) 4x2 - 4x + 5 ≤ 0
<=> 4x2 - 4x + 1 + 4 ≤ 0
<=> ( 2x - 1 )2 + 4 ≤ 0
<=> ( 2x - 1 )2 ≤ -4 ( vô lí )
=> BPT vô nghiệm ( đpcm )
f) x2 + x + 1 ≤ 0
<=> x2 + 2.1/2.x + 1/4 + 3/4 ≤ 0
<=> ( x + 1/2 )2 + 3/4 ≤ 0
<=> ( x + 1/2 )2 ≤ -3/4 ( vô lí )
=> BPT vô nghiệm ( đpcm )
a,Ta có :\(x^2+2x+2=\left(x^2+2x+1\right)+1\)
\(=\left(x+1\right)^2+1\)
Do \(\left(x+1\right)^2\ge0< =>\left(x+1\right)^2+1\ge1\)
=> BPT vô nghiệm
b,Ta có :\(4x^2-4x+5=\left[\left(2x\right)^2-2.2x+1\right]+4\)
\(=\left(2x-1\right)^2+4\)
Do \(\left(2x-1\right)^2\ge0< =>\left(2x-1\right)^2+4\ge4\)
=> BPT vô nghiệm
c,Ta có :\(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2+2.\frac{1}{2}.x+\frac{1}{2}^2\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Do \(\left(x+\frac{1}{2}\right)^2\ge0< =>\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> BPT vô nghiệm
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
a) ( x +2 )2 - ( 3x - 1 ) ( x +2 ) = 0
<=> (x+2)(x+2-3x+1) = 0
<=> (x+2)(-2x+3) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\-2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{3}{2}\end{cases}}}\)
b) ( 2x - 1 )2 - ( 2x + 5 ) ( 2x - 5 ) = 18
<=> 4x2 -4x +1 - (4x2-25) =18
<=> 4x2 -4x +1 - 4x2 + 25 = 18
<=> - 4x + 26 = 18
<=> - 4x = 18 - 26
<=> - 4x = -8
<=> x = 2
c) ( 2x + 3 )2 - ( x - 5 )2 = 0
<=> [( 2x + 3 ) - ( x - 5 )].[( 2x + 3 ) + ( x - 5 )] = 0
<=> (2x +3 -x +5) . (2x +3 + x - 5) = 0
<=> (x +8)(3x-2) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+8=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-8\\x=\frac{2}{3}\end{cases}}}\)
d) 5x3 + 3x - 8 = 0
<=> (5x3 -5x) +(8x-8) = 0
<=> 5x(x2 - 1) + 8(x-1) = 0
<=> 5x(x - 1)(x+1) + 8(x-1) = 0
<=> (x - 1)[5x(x+1) + 8] = 0
<=> (x-1)(5x2+5x +8 ) = 0
<=> (x-1).5.(x2+x+8/5) = 0
<=> 5.(x-1)(x2+x+1/4 + 27/20) = 0
\(\Leftrightarrow\left(x-1\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{27}{20}\right]\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\\left(x+\frac{1}{2}\right)^2+\frac{27}{20}=0\end{cases}\Leftrightarrow x=1}\)vỉ \(\left(x+\frac{1}{2}\right)^2+\frac{27}{20}>0\)với mọi x
Vậy x = 1
\(2x\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\left(2x+1\right)\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)
\(9\left(3x-2\right)-x\left(2-3x\right)=0\)
\(9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\left(9+x\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)
\(\left(2x-1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
a) 16x^2 - (4x - 5)^2 = 15
<=> 16x^2 - 16x^2 + 40x - 25 = 15
<=> 40x = 40
<=> x = 1
b) (2x + 3)^2 - 4(x - 1)(x + 1) = 49
<=> 4x^2 + 12x + 9 - 4x^2 - 4x + 4x + 4 = 49
<=> 12x + 13 = 49
<=> 12x = 36
<=> x = 3
c) (2x + 1)(1 - 2x) + (1 - 2x)^2 = 18
<=> 1 - 4x^2 + 1 - 4x + 4x^2 = 18
<=> 2 - 4x = 18
<=> -4x = 16
<=> x = -4
d)2(x + 1)^2 - (x - 3)(x + 3) - (x - 4)^2 = 0
<=> 2x^2 + 4x + 2 - x^2 + 3^2 - x^2 + 8x - 16 = 0
<=> 12x - 5 = 0
<=> 12x = 5
<=> x = 5/12
e) (x - 5)^2 - x(x - 4) = 9
<=> x^2 - 10x + 25 - x^2 + 4x = 9
<=> -6x + 25 = 9
<=> -6x = 9 - 25
<=> -6x = -16
<=> x = -16/-6 = 8/3
f) (x - 5)^2 + (x - 4)(1 - x) = 0
<=> x^2 - 10x + 25 + x - x^2 - x - 4 + 4x = 0
<=> -5x + 21 = 0
<=> -5x = -21
<=> x = 21/5