Tìm x biết :
25+x=(-3456÷3+678×1250)
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3:
a: -5x-16=-46
=>-5x=-46+16=-30
=>5x=30
=>\(x=\dfrac{30}{5}=6\)
b: -80-(-16-x)=20
=>\(-x-16=-80-20=-100\)
=>x+16=100
=>x=100-16=84
2:
a: \(\left(45-3456\right)+3456\)
\(=45+3456-3456\)
=45
b: \(\left(77+22-75\right)-\left(-25-78-13\right)\)
\(=77+13+22+78-75+25\)
\(=90+100-50\)
=140
c: \(-25\cdot63-25\cdot37\)
\(=-25\left(63+37\right)\)
\(=-25\cdot100=-2500\)
678 x 91 - 678 xa = 678
678 x ( 91-a ) =678
91 -a = 6780:678
91-a = 10
a = 91-10
a = 81
vậy a=81
678*91 -678*a = 6780 thi a = 81 vi 91-81= 10 ma 10*678 =6780 vay thoai man dau bai
Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)
\(\left(y+\dfrac{2}{5}\right)^{100}\ge0\forall y\)
\(\left(z-3\right)^{678}\ge0\forall z\)
Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+\dfrac{2}{5}\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)
Dấu '=' xảy ra khi \(\left(x,y,z\right)=\left(\dfrac{1}{5};\dfrac{-2}{5};3\right)\)
Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0,\left(y+0,4\right)^{100}\ge0,\left(z-3\right)^{678}\ge0\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0,4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-0,4\\z=3\end{matrix}\right.\)
Vậy \(\left(x,y,z\right)=\left(\dfrac{1}{5};-0,4;3\right)\)
Vì \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)
\(\left(y+0,4\right)^{100}\ge\forall y\)
\(\left(z-3\right)^{678}\ge0\forall z\)
\(\Rightarrow\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}\ge0\)
mà \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0,4\right)^{100}+\left(z-3\right)^{678}=0\)
Dấu ''='' xảy ra khi \(x=\dfrac{1}{5};y=-0,4;z=3\)
25+x=846348
=>x=846348-25
x=846323
ok
xin 5 tích
25 + x = ( -3456 : 3 + 678 x 1250 )
25 + x = 846348
x = 846348 - 25
x = 846323