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Bài 1 : Tính :
\(a)49-\left(-54\right)-23=103-23\)
\(=80\)
\(b)\left(-25\right).68+\left(-34\right).\left(-250\right)=-1700+8500\)
\(=6800\)
\(c)1999+\left(-2000\right)+2001+\left(-2002\right)=\left(-1\right)+\left(-1\right)\)
\(=-2\)
\(d)515+\left[72+\left(-515\right)+\left(-32\right)\right]=515+\left(-475\right)\)
\(=40\)
\(e)\left(2736-75\right)-2736+175=2661-2736+175\)
\(=100\)
\(g)-2020-\left(157-2020\right)-\left(-257\right)=-2020-\left(-1863\right)-\left(-257\right)\)
\(=100\)
1,
a,\(2020-\left(249+2020\right)+\left(249-573\right)\)
\(=2020-249-2020+249-573\)
\(=-573\)
b,\(\left|-257\right|+\left(-3\right)^0-\left(18+257\right)\)
\(=257+1-18-257\)
\(=1-18=-17\)
\(c,25.\left(85-47\right)-85.\left(47+25\right)\)
\(=25.85-25.47-47.85+85.25\)
\(=85.\left(25-47+25\right)-25.47\)
\(=85.3-25.47\)
\(=-920\)
2,
\(a,15-5.\left(x+2\right)=-30\)
\(=>5.\left(x+2\right)=15+30=45\)
\(=>x+2=\frac{45}{5}=9\)
\(=>x=7\)
\(b,\left(x+2\right)^2+5=105\)
\(=>\left(x+2\right)^2=100\)
\(=>\left(x+2\right)^2=10^2\)
\(=>x+2=10\)
\(=>x=8\)
\(c,\left|2x-5\right|-\left(-6\right)=11\)
\(=>\left|2x-5\right|=11-6=5\)
\(=>\orbr{\begin{cases}2x-5=5\\2x-5=-5\end{cases}}\)
\(=>\orbr{\begin{cases}2x=5-5=0\\2x=-5+5=0\end{cases}=>x=0}\)
Câu 1:
a)\(1999+\left(-2000\right)+2001+\left(-2002\right)=1999-2000+2001-2002=-2\)
b)\(49-\left(-54\right)-23=49+54-23=80\)
c)\(\left(-25\right).68+\left(-34\right).\left(-250\right)=-1700+8500=6800\)
Câu 2:
a)x+7=3
\(\Rightarrow x=4\) Vậy x=4
b)\(\left|x\right|=7\Rightarrow x=\left\{\pm7\right\}\)mà x<0 nên x=-7 thì thỏa mãn đề bài
c) hình như thiếu điều kiện
1. Tính:
a) 1999 + ( - 2000 ) + 2001 + ( - 2002 )
= ( 1999 + 2001 ) + [ ( -2000 ) + ( -2002 ) ]
= 4000 + ( -4002 )
= -2
b) 49 - ( - 54 ) - 23
= 49 + 54 - 23
= ( 49 - 23 ) + 54
= 26 + 54
= 80
c) ( - 25 ) . 68 + ( -34 ) . ( - 250 )
= 1700 + 8500
= 6800
d) Tính tổng các giá trị x thuộc Z, biết: -3 < | x | < 4
=> | x | = 0, 1, 2; 3
=> | x | thuộc { ( - 3 ) + ( - 2 ) + ( -1) + 0 + 1 + 2 + 3 = 0
2. Tìm x thuộc Z
a) x + 7 = 3
=> x = 3 - 7
=> x = -4
b) | x | = 7 và x < 0
=> x = -7; 7
Mà x < 0
=> x = -7
c) | x | > x
=> | x | > x <=> x thuộc Z trừ
Bài 1:
a) x + 60
b) -p
Bài 2:
a) = 2736 - 2736 - 75
= - 75
b) = - 2002 - 57 + 2002
= - 2002 + 2002 - 57
= - 57
a) x + 22 + (-14) + 52 = x + [ 22 + 52 ] + (-14) = x + 74 + (-14)
= x + [ 74 + (-14) ] = x + 60
b) (-90) - (p + 10) + 100 = (-90) - p - 10 + 100 = - p - 90 - 10 + 100
= -p - (90 + 10) + 100 = -p - 100 + 100 = -p + [ (-100) + 100] = -p
Bài 1. Tính hợp lý ( nếu có thể ) :
a) ( 509 - 42 ) - ( 509 + 158 )
= 509 - 42 - 509 - 158
=(509 - 509) - ( 42+ 158)
= 0 - 200
= -200
b) - ( 510 - 427 ) - 473 + 510
= -510 +427 -473 +510
= (510-510) + (427-473)
= 0 + (-46)
=-46
c) 1995 - (-2021 + 1994) - 1
= 1995 +2021 -1994 -1
=( 1995 -1994) + (2021-1)
= 1+2020
= 2021
d) 2020 - [ - 1079 -(-1179 + 3020 )]
= 2020 - [ -1079 +1179-3020]
= 2020 +1079 -1179 +3020
=(2020+3020) + (1079 - 1179)
= 5040 +(-100)
=4940
e) 12+(-47).12-6.(-12)
= 12.1 + (-47) .12 -(-6).12
= 12.1 -47 .12 +6 .12
= 12. (1-47+6)
=12.(-40)
=-480
g) 32.(132-247)-132.(32-247)
=32.132 -32.247 - (132.32 - 132.247)
= 32.132 -32.247 - 132.32 + 132.247
= ( 32.132 - 132.32) + ( 132.247 - 32.247)
= 0 + 247.(132-32)
= 247.100
= 24700
i) (-25) .68+(-34).(-250)
= ( -25) . ( -2) . ( -34) + ( -34) . ( -250)
= 50 . (-34) + ( -34) . (-250)
= (-34) . [ 50 + ( - 250) ]
= ( - 34 ) . ( -200)
= 6800
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}