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6. (x - 2)+ (-2) = 20 - 4x
\(6x-12+\left(-2\right)=20-4x\)
\(6x+4x=20+12+2\)
\(10x=34\)
\(\Rightarrow x=34\div10\)
\(\Rightarrow x=3,4\)
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6 . (x - 2) + (-2) = 20 - 4x
6 . (x - 2) + (-2) + 4x = 20
6x - (6 . 2) + 4x + (-2) = 20
6x - (6 . 2) + 4x = 20 - 2
6x - (6 . 2) + 4x = 18
(6x + 4x) - (6 . 2) = 18
(6x + 4x) - 12 = 18
(6x + 4x) = 18 + 12
(6x + 4x) = 30
10x = 30
x = 30 : 10
x = 3.
Vậy x = 3.
\(\left(x-7\right)\left(x+2019\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-7=0\\x+2019=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-2019\end{cases}}\)
\(9-25=\left(7-x\right)-\left(25+7\right)\)
\(\Leftrightarrow-16=7-x-25-7\)
\(\Leftrightarrow-x=-16+25\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
\(2\left(4x-2x\right)-7x=15\)
\(\Leftrightarrow4x-7x=15\)
\(\Leftrightarrow x=-5\)
a ) 9 - 25 = ( 7 - x ) - ( 25 + 7 )
9 - 25 = 7 - x - 25 - 7
9 - 25 - 7 + 25 + 7 = -x
9 = - x
=> x = -9
Vậy x = -9
b) 2 . ( 4x - 2x ) - 7x = 15
8x - 4x - 7x = 15
-3x = 15
x = 15 : ( - 3 )
x = -5
Vậy x = -5
c ) ( x - 7 ). ( x + 2019 ) = 0
=> x - 7 = 0 hoặc x + 2019 = 0
=> x = 7 hoặc x = - 2019
vậy x \(\in\){ 7 ; -2019 }
Có: \(\dfrac{2}{x-2}-\dfrac{2}{x+2}=2\left(dkxd:x\ne\pm2\right)\)
\(\Rightarrow2\cdot\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)=2\)
\(\Rightarrow\dfrac{1}{x-2}-\dfrac{1}{x+2}=1\)
\(\Rightarrow\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=1\)
\(\Rightarrow\dfrac{x+2-x+2}{x^2-4}=1\)
\(\Rightarrow\dfrac{4}{x^2-4}=1\)
\(\Rightarrow x^2-4=4\)
\(\Rightarrow x^2=8\)
Thay \(x^2=8\) vào \(\left(x^2+1\right)^2\), ta được:
\(\left(8+1\right)^2=9^2=81\)
\(\dfrac{2}{x-2}\) - \(\dfrac{2}{x+2}\) - 2 = 0
2.(\(\dfrac{1}{x-2}\) - \(\dfrac{1}{x+2}\) - 1) = 0
\(\dfrac{1}{x-2}\) - \(\dfrac{1}{x+2}\) - 1 = 0
\(\dfrac{x+2-\left(x-2\right)-\left(x-2\right).\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\) = 0
\(x\) + 2 - \(x\) + 2 - (\(x^2\) + 2\(x\) - 2\(x\) - 4) = 0
4 - \(x^2\) + 4 = 0
8 - \(x^2\) = 0
\(x^2\) = 8
Thay \(x^2\) = 8 vào ( \(x^2\) + 1)2 ta có: (\(x^2\) + 1) = (8 + 1)2 = 92 = 81
\(90=2\cdot3^2\cdot5\\ 126=2\cdot3^2\cdot7\\ ƯCLN\left(90,126\right)=2\cdot3^2=18\\ ƯC\left(90,126\right)=Ư\left(18\right)=\left\{1;2;3;6;9;18\right\}\)
\(A=x^2-4x+5\)
\(A=x^2-2\cdot2x+2^2-2^2+5\)
\(A=\left(x-2\right)^2-2^2+5\)
\(A=\left(x-2\right)^2-4+5\)
\(A=\left(x-2\right)^2+1\)
Ta có : \(\left(x-2\right)^2\ge0\)
\(\Rightarrow\left(x-2\right)^2+1\ge1\)
=> GTNN của x^2 -4x +5 là 1
Thanks !