tìm x
x3 - 2x2 - x + 2 = 0
gấp ạ đầy đủ tick ạ
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(2,5 x 11,25 + 11,25 x 7,5) x( 75 x 11 - 75 : 0,1 - 75)
ai nhanh tớ sẽ tick ạ, ghi đầy đủ lời giải ạ= ( 11,25 x (2,5 + 7,5) ) x ( 75 x 11 - 75 : 0,1 -75 x1)
=( 11,25 x 10) x (75 x ( 11 -1 :0,1 ) )
= 112,5 x ( 75 x 100 )
= 112,5 x 7500
= 1125000
( 2,5 \(\times\) 11,25 \(\) + 11,25 \(\times\)7,5) \(\times\)( 75 \(\times\) 11 - 75: 0,1 - 75)
= (2,5 \(\times\) 11,25 + 11,25 \(\times\) 7,5) \(\times\)( 75 \(\times\) 11 - 75 \(\times\) 10 - 75 \(\times\) 1)
=( 2,5 \(\times\) 11,25 + 11,25 \(\times\) 7,5) \(\times\) 75 \(\times\)( 11 - 10 - 1)
= ( 2,5 \(\times\) 11,25 + 11,25 \(\times\)7,5 ) \(\times\) 75 \(\times\) 0
= 0
Đặt A = x + 1 + x + 2 + x + 3 + ....+ x + 20 = 25 x
Số số hạng của dãy: (20 - 1) + 1 = 20
=> A = 20 x + (1+ 20).20 : 2 = 25 x => 210 = 5 x => x = 42
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(x^3+x^2-9x-9=0\Leftrightarrow x^2\left(x+1\right)-9\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=-1;\pm3\)
c, \(x^2-3x-10=0\Leftrightarrow x^2+2x-5x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=5;-2\)
\(\dfrac{6}{x}+\dfrac{1}{2}=2\\ \dfrac{6}{x}=2-\dfrac{1}{2}\\ \dfrac{6}{x}=\dfrac{4}{2}-\dfrac{1}{2}\\ \dfrac{6}{x}=\dfrac{3}{2}\\ x=6:\dfrac{3}{2}\\ x=\dfrac{6x2}{3}\\ x=4\)
Đủ chi tiết chưa nhỉ ??
a,\(x^2y-4y=y\left(x^2-4\right)=y\left(x-2\right)\left(x+2\right)\)
b,\(x^2-y^2-2x+1=\left(x^2-2x+1\right)-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x-y-1\right)\)
c,\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(5x-1\right)\left(x+y\right)\)
x2y - 4y = y( x2 - 4 ) = y( x - 2 )( x + 2 )
x2 - y2 - 2x + 1 = ( x2 - 2x + 1 ) - y2 = ( x - 1 )2 - y2 = ( x - 1 - y )( x - 1 + y )
5x2 + 5xy - x - y = ( 5x2 + 5xy ) - ( x + y ) = 5x( x + y ) - ( x + y ) = ( x + y )( 5x - 1 )
1/ went
2/ prepared
3/ took
4/ played
5/ watched
6/ danced
7/ rained
Lời giải:
ĐKXĐ: $x\geq 0$
$-\sqrt{x}(\sqrt{x}-1)>0$
$\Leftrightarrow \sqrt{x}(\sqrt{x}-1)<0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} \sqrt{x}>0\\ \sqrt{x}-1<0\end{matrix}\right.\\ \left\{\begin{matrix} \sqrt{x}<0\\ \sqrt{x}-1>0\end{matrix}\right. (\text{TH này hiển nhiên vô lý})\end{matrix}\right. \)
\(\Leftrightarrow \left\{\begin{matrix} x>0\\ 0\leq x< 1\end{matrix}\right.\Leftrightarrow 0< x< 1\)
\(x^3-2x^2-x+2=0\Leftrightarrow x^3-x-2x^2+2=0\)
\(\Leftrightarrow x\left(x^2-1\right)-2\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=2;\pm1\)
\(x^3-2x^2-x+2=0\)
\(x^2.\left(x-2\right)-\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^2-1\right)=0\)
\(\left(x-2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-2=0\\x-1=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=1\\x=-1\end{cases}}}\)
Vậy \(x=2;x=1;x=-1\)