tìm x, biết :
2.(x+7)-(2x+3).(x-1)-8= 6x
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a) \(-\frac{7}{2}:\left(3-x\right)-0,75=\frac{1}{4}\)
\(\Leftrightarrow\frac{-7}{2\left(3-x\right)}=\frac{1}{4}+0,75=0,25+0,75=1\)
\(\Leftrightarrow2\left(3-x\right)=-7\)
\(\Leftrightarrow3-x=-\frac{7}{2}\)
\(\Leftrightarrow x=3+\frac{7}{2}\)
\(\Leftrightarrow x=\frac{13}{2}\)
b) \(\left|2x-\frac{4}{3}\right|-1\frac{1}{3}=-\frac{8}{9}\)
\(\Leftrightarrow\left|2x-\frac{4}{3}\right|=\frac{4}{3}-\frac{8}{9}=\frac{4}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{4}{3}=\frac{4}{9}\\2x-\frac{4}{3}=-\frac{4}{9}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{4}{3}+\frac{4}{9}=\frac{16}{9}\\2x=\frac{4}{3}-\frac{4}{9}=\frac{8}{9}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{9}\\x=\frac{4}{9}\end{cases}}\)
\(\left(2x+1\right)\left(x^2-x\right)+x\left(5+x-2x^2\right)=3x+7\)
\(2x^3-2x^2+x^2-x+5x+x^2-2x^3=3x+7\)
\(5x-x=3x+7\)
\(4x-3x=7\)
\(x=7\)
(2x+1)(x^2-x)+x(-2x^2+x+5)=3x+7
=>2x^3-2x^2+x^2-x-2x^3+x^2+5x=3x+7
=>-x^2-x+x^2+5x=3x+7
=>4x=3x+7
=>x=7
Vì |x-1|+|x+1| luôn ko âm.
Với x âm .
=>2x-3 âm(loại)
Với x=1.
=?2x-3 âm (loại)
=>x>1.
=>|x-1|+|x+1|=x-1+x+1=2x=2x+3.
Hơi vô lí nhỉ!
Ta có: |x - 1| + |x + 1| = 2x - 3
Vì \(\hept{\begin{cases}\left|x-1\right|\ge0\\\left|x+1\right|\ge0\end{cases}}\) \(\Rightarrow\) \(\left|x-1\right|+\left|x+1\right|\ge0\)\(\Rightarrow2x-3\ge0\)\(\Rightarrow2x\ge3\Rightarrow x\ge\frac{3}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1+x+1=2x-3\\x-1+x+1=-\left(2x-3\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=2x-3\\2x=-2x+3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2x=-3\\2x+2x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}0x=-3\\4x=3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(loai\right)\\x=\frac{3}{4}9\left(loai\right)\end{cases}}}\)
Vậy không có giá trị x thỏa mãn.
Tìm x biết :
45 - [ ( 72 - 8 . x ) : 4 + 7 ] . 3 = 0
[ ( 72 - 8 . x ) :4 + 7 ] . 3 = 45 - 0
[ ( 72 - 8 . x) : 4 + 7 ] . 3 = 45
( 72 - 8 . x ) : 4 + 7 = 45 : 3
( 72 - 8 . x ) : 4 + 7 = 15
( 72 - 8 . x ) : 4 = 15 - 7
( 72 - 8 . x ) : 4 = 8
72 - 8 . x = 8 × 4
72 - 8 . x = 32
8 . x = 72 - 32
8 . x = 40
x = 40 : 8
x = 5
Vậy x = 5
\([\left(72-8\cdot x\right):4+7]\cdot3=45-0\)
\([\left(72-8\cdot x\right):4+7]\cdot3=45\)
\(\left(72-8\cdot x\right):4+7=45:3\)
\(\left(72-8\cdot x\right):4+7=15\)
\(\left(72-8\cdot x\right):4=15-7\)
\(\left(72-8\cdot x\right):4=8\)
\(72-8\cdot x=8\cdot4\)
\(72-8\cdot x=32\)
\(8x=72-32\)
\(8x=40\)
\(x=40:8\)
\(x=5\)
( x - 1 )( x + 2 ) - x - 2 = 0
<=> ( x - 1 )( x + 2 ) - ( x + 2 ) = 0
<=> ( x + 2 )( x - 2 ) = 0
<=> x = ±2
( 2x - 7 )3 = 8( 7 - 2x )2
<=> ( 2x - 7 )3 - 8( 2x - 7 )2 = 0
<=> ( 2x - 7 )2( 2x - 15 ) = 0
<=> x = 7/2 hoặc x = 15/2
2(x + 7) - (2x + 3).(x - 1) - 8 = 6x
<=> (2x + 14) - (2x + 3)(x - 1) - 8 - 6x = 0
<=> 2x + 14 - (2x2 + 3x - 2x - 3) - 8 - 6x = 0
<=> 2x + 14 - (2x2 + x - 3) - 8 - 6x = 0
<=> 2x + 14 - 2x2 - x + 3 - 8 - 6x = 0
<=> -2x2 - 5x + 6 = 0
<=> 2x2 + 5x - 6 = 0
<=> \(x^2+\frac{5}{2}x-3=0\)
\(\Leftrightarrow x^2+2.x.\frac{5}{4}+\frac{25}{16}-\frac{73}{16}=0\)
\(\Leftrightarrow x^2+2.x.\frac{5}{4}+\frac{25}{16}=\frac{73}{16}\)
\(\Leftrightarrow\left(x+\frac{5}{4}\right)^2=\frac{73}{16}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{5}{4}=\frac{73}{16}\\x+\frac{5}{4}=-\frac{73}{16}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{73}{16}-\frac{5}{4}\\x=-\frac{73}{16}-\frac{5}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{53}{16}\\x=-\frac{93}{16}\end{cases}}\)
2(x+7)-(2x+3)(x-1)-8=6x
\(\Leftrightarrow2x+14-2x^2+2x-3x+3-8=6x\)
\(\Leftrightarrow\) \(-2x^2+2x+2x-3x+3-8+14=6x\)
\(\Leftrightarrow-2x^2+x+9=6x\)
\(\Leftrightarrow-2x^2-5x+9=0\)
\(\Leftrightarrow\left(x-\left(\frac{-5+\sqrt{97}}{4}\right)\right)\left(x+\left(\frac{-5-\sqrt{97}}{4}\right)\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{-5+\sqrt{97}}{4}=0\\x+\frac{-5-\sqrt{97}}{4}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+97}{4}\\x=\frac{5-\sqrt{97}}{4}\end{cases}}\)