Question

bài 1:tính

a) (-25).68 + (-34).(-250) b) 1999+(-2000)+2001+(-2002)

c)515+[72+(-515)+(-32)] d) (2736-75)-2736+175

e) -2020 - (157-2020)-(-257)

bài 2: tìm các số nguyên x biết

a) x-|-2| =|-18| b)2x - |+14|=|-14|

c) |x+4| + 5= 20-(-12-7) d)15-|2-x|=(-2)²

e) |15-x| + |-25| =|-55| g) |17-(-4)|+|-24-(-5)|=||-x+3|

Answer 1

Bài 1: Tính

a) Ta có: \(\left(-25\right)\cdot68+\left(-34\right)\cdot\left(-250\right)\)

\(=-25\cdot68+\left(-340\right)\cdot\left(-25\right)\)

\(=-25\cdot\left(68-340\right)\)

\(=-25\cdot\left(-272\right)\)

\(=6800\)

b) Ta có: \(1999+\left(-2000\right)+2001+\left(-2002\right)\)

\(=1999-2000+2001-2002\)

\(=-1-1=-2\)

c) Ta có: \(515+\left[72+\left(-515\right)+\left(-32\right)\right]\)

\(=515+72-515-32\)

\(=40\)

d) Ta có: \(\left(2736-75\right)-2736+175\)

\(=2736-75-2736+175\)

\(=100\)

e) Ta có: \(-2020-\left(157-2020\right)-\left(-257\right)\)

\(=-2020-157+2020+257\)

\(=100\)

Bài 2: Tìm x

a) Ta có: \(x-\left|-2\right|=\left|-18\right|\)

\(\Leftrightarrow x-2=18\)

hay x=20

Vậy: x=20

b) Ta có: \(2x-\left|+14\right|=\left|-14\right|\)

\(\Leftrightarrow2x-14=14\)

\(\Leftrightarrow2x=28\)

hay x=14

Vậy: x=14

c) Ta có: \(\left|x+4\right|+5=20-\left(-12-7\right)\)

\(\Leftrightarrow\left|x+4\right|+5=20+12+7\)

\(\Leftrightarrow\left|x+4\right|=39-5=34\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=34\\x+4=-34\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-38\end{matrix}\right.\)

Vậy: x∈{30;-38}

d) Ta có: \(15-\left|2-x\right|=\left(-2\right)^2\)

\(\Leftrightarrow15-\left|2-x\right|=4\)

\(\Leftrightarrow\left|2-x\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2-x=11\\2-x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=13\end{matrix}\right.\)

Vậy: x∈{-9;13}

e) Ta có: \(\left|15-x\right|+\left|-25\right|=\left|-55\right|\)

\(\Leftrightarrow\left|15-x\right|+25=55\)

\(\Leftrightarrow\left|15-x\right|=30\)

\(\Leftrightarrow\left[{}\begin{matrix}15-x=30\\15-x=-30\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-15\\x=45\end{matrix}\right.\)

Vậy: x∈{-15;45}

g) Ta có: \(\left|17-\left(-4\right)\right|+\left|-24-\left(-5\right)\right|=\left|-x+3\right|\)

\(\Leftrightarrow\left|17+4\right|+\left|-24+5\right|=\left|3-x\right|\)

\(\Leftrightarrow\left|3-x\right|=40\)

\(\Leftrightarrow\left[{}\begin{matrix}3-x=40\\3-x=-40\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-37\\x=43\end{matrix}\right.\)

Vậy: x∈{-37;43}