X = (70-6):8 

X = 8

70 : x = 8 (dư 6)

       x = (70 - 6) : 8

       x = 8

x=8

a) Ta có : \(C=15-\left(|x+3|+|y+2|\right)\)

Vì \(|x+3|\ge0\)\(\forall x\)

    \(|y+2|\ge0\)\(\forall y\)

\(\Rightarrow|x+3|+|y+2|\ge0\)\(\forall x,y\)

\(\Rightarrow15-\left(|x+3|+|y+2|\right)\le15\)\(\forall x,y\)

hay \(C\le15\)

\(\Rightarrow maxC=15\Leftrightarrow x+3=0\)và \(y+2=0\)

                               \(\Leftrightarrow x=-3\)và \(y=-2\)

Vậy GTLN của C là 15 \(\Leftrightarrow x=-3\)và \(y=-2\)

b) Ta có : \(D=70-[|x-2018|+\left(y-3\right)^2]\)

Vì \(|x-2018|\ge0\)\(\forall x\)

     \(\left(y-3\right)^2\ge0\)\(\forall y\)

\(\Rightarrow|x-2018|+\left(y-3\right)^2\ge0\)\(\forall x,y\)

\(\Rightarrow D\le70\)

\(\Rightarrow maxD=70\Leftrightarrow x-2018=0\)và \(\left(y-3\right)^2=0\) 

                               \(\Leftrightarrow x=2018\)và \(y-3=0\)

                               \(\Leftrightarrow x=2018\)và \(y=3\)

Vậy max D = 70 \(\Leftrightarrow\)x = 2018 và y = 3

Bài 1: 

a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{1}{2}\\x-\dfrac{1}{3}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=-\dfrac{1}{6}\end{matrix}\right.\)

\(a.\dfrac{4}{5}+x=\dfrac{8}{10}\\ \Rightarrow x=\dfrac{8}{10}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{8}{10}-\dfrac{8}{10}\\ \Rightarrow x=0\\ b.\dfrac{1}{7}-7\left(\dfrac{2}{3}-x\right)=-\dfrac{4}{7}\\ \Rightarrow\dfrac{1}{7}-\dfrac{14}{3}+7x=-\dfrac{4}{7}\\ \Rightarrow-\dfrac{95}{21}+7x=-\dfrac{4}{7}\\\Rightarrow7x=\dfrac{83}{21}\\ \Rightarrow x=\dfrac{83}{147} \)

`a)4/5+x=8/10`

`=>x=8/10-4/5`

`=>x=4/5-4/5`

`=>x=0`

Vậy `x=0`

__

`b)1/7-7(2/3-x)=-4/7`

`=>7(2/3-x)=1/7-(-4/7)`

`=>7(2/3-x)=1/7+4/7`

`=>7(2/3-x)=5/7`

`=>2/3-x=5/7:7`

`=>2/3-x=5/7xx1/7`

`=>2/3-x=5/49`

`=>x=2/3-5/49`

`=>x=98/147-15/147`

`=>x=83/147`

Vậy `x=83/147`

a: \(x^8+x^4+1\)

\(=x^8+2x^4+1-x^4\)

\(=\left(x^4+1\right)^2-x^4\)

\(=\left(x^4+1+x^2\right)\left(x^4+1-x^2\right)\)

\(=\left(x^4+2x^2+1-x^2\right)\left(x^4-x^2+1\right)\)

\(=\left(x^4-x^2+1\right)\cdot\left[\left(x^2+1\right)^2-x^2\right]\)

\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)

b: \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2\)

\(=\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2\)

\(=\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+1+x\right)\)

\(=\left(x^2+x+1\right)\left(x^2+2x+1\right)\)

\(=\left(x^2+x+1\right)\left(x+1\right)^2\)

nhanh nha mấy bn

a) 56

b) 20

c) khó quá mình ko biết

nho k minh nhe