X x 12,5 x X x 10 = 36,4
X lớn là x nhé còn x nhỏ là dấu nhân
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a. 12,5 - \(x\) = \(x\) + 12
\(x\) + 12 + \(x\) = 12,5
\(x\) x 1 + \(x\) x 1 = 12,5 - 12
\(x\) x (1 + 1) = 0,5
\(x\) x 2 = 0,5
\(x\) = 0,5 : 2
\(x\) = 0,25
b, 8 \(\times\) \(x\) + 3 \(\times\) \(x\) + 2 \(\times\) \(x\) = 20
\(x\) \(\times\) (8 + 3 + 2) =20
\(x\) \(\times\) 13 = 20
\(x\) = 20 : 13
\(x\) = \(\dfrac{20}{13}\)
\(\frac{1}{2}x+\frac{2}{4}x=-\frac{18}{25}.\)
\(\Leftrightarrow\frac{1}{2}x+\frac{1}{2}x=-\frac{18}{25}\)
\(\Leftrightarrow\left(\frac{1}{2}+\frac{1}{2}\right)x=-\frac{18}{25}\)
\(\Leftrightarrow x=-\frac{18}{25}\)
\(5\cdot x+10\cdot9=990\)
\(\Rightarrow5\cdot x+90=990\)
\(\Rightarrow5\cdot x=900\)
\(\Rightarrow x=\dfrac{900}{5}\)
\(\Rightarrow x=180\)
_____________
\(1045:\left[215-\left(3\cdot x-24\right)\right]=5\)
\(\Rightarrow215-\left(3\cdot x-24\right)=1045:5\)
\(\Rightarrow215-\left(3\cdot x-24\right)=209\)
\(\Rightarrow3\cdot x-24=215-209\)
\(\Rightarrow3\cdot x-24=6\)
\(\Rightarrow3\cdot x=30\)
\(\Rightarrow x=10\)
`@` `\text {Ans}`
`\downarrow`
`5.x + 10.9 = 990`
`\Rightarrow 5x + 90 = 990`
`\Rightarrow 5x = 990 - 90`
`\Rightarrow 5x = 900`
`\Rightarrow x = 900 \div 5`
`\Rightarrow x = 180`
Vậy, `x = 180`
\(1045 \div [ 215 - (3 . x - 24 ) ] = 5\)
`\Rightarrow 1045 \div (215 - 3x + 24) = 5`
`\Rightarrow 191 + 3x = 1045 \div 5`
`\Rightarrow 191 + 3x = 209`
`\Rightarrow 3x = 209 - 191`
`\Rightarrow 3x =18`
`\Rightarrow x = 18 \div 3`
`\Rightarrow x = 6`
Vậy, `x = 6.`
x*12,5 -x +x *2,75=1050
x*12,5 -x*1+x*2,75=`1050
x*(12,5-1+2,75)=1050
x*14,25=1050
x=1050:14,25
x= 73,68421053
a/ => 4x2 - 4x + 1 + 4x2 + 4x + 1 = 16
=> 8x2 = 14
=> x2 = 14/8
=> x = \(\frac{\sqrt{7}}{2}\) hoặc x = \(-\frac{\sqrt{7}}{2}\)
b/ => 6x2 - (6x2 - 11x - 10) = 17
=> 6x2 - 6x2 + 11x + 10 = 17
=> 11x = 7
=> x = 7/11
c/ => 2x(x + 5) - x2 - 5x = 0
=> 2x(x + 5) - x(x + 5) = 0
=> (x + 5)(2x - x) = 0
=> x(x + 5) = 0
=> x = 0
hoặc x + 5 = 0 => x = -5
Vậy x = 0 ; x = -5
d/ \(x^2+\frac{1}{x^2}+2x+\frac{2}{x}=-3\)
đề là như vầy hả
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)
\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)
\(\Leftrightarrow x=122\)
\(x\left(x+y+z\right)=10\) (1)
\(y\left(y+z+x\right)=25\) (2)
\(z\left(z+x+y\right)=-10\) (3)
Lấy (1) + (2) + (3) theo vế ta có:
\(x\left(x+y+z\right)+y\left(y+z+x\right)+z\left(z+x+y\right)=10+25-10\)
\(\Leftrightarrow\)\(\left(x+y+z\right)^2=25\)
\(\Leftrightarrow\)\(x+y+z=\pm\sqrt{25}=\pm5\)
Nếu \(x+y+z=5\) thì: \(\hept{\begin{cases}x=2\\y=5\\z=-2\end{cases}}\)
Nếu \(x+y+z=-5\)thì \(\hept{\begin{cases}x=-2\\y=-5\\z=2\end{cases}}\)
Vậy...
\(x\) x 12,5 x \(x\) x 10 = 36,4
=> \(x\)2 x 12,5 x 10 = 36,4
=> \(x\)2 x 125 = 36,4
=> \(x\)2 = 36,4 : 125
=> \(x\)2 = 0,2912
=> \(x\)= \(\sqrt{0,2912}\) hoặc \(-\sqrt{0,2912}\)