Bài 1 : Tính :

\(a)49-\left(-54\right)-23=103-23\)

\(=80\)

\(b)\left(-25\right).68+\left(-34\right).\left(-250\right)=-1700+8500\)

\(=6800\)

\(c)1999+\left(-2000\right)+2001+\left(-2002\right)=\left(-1\right)+\left(-1\right)\)

\(=-2\)

\(d)515+\left[72+\left(-515\right)+\left(-32\right)\right]=515+\left(-475\right)\)

\(=40\)

\(e)\left(2736-75\right)-2736+175=2661-2736+175\)

\(=100\)

\(g)-2020-\left(157-2020\right)-\left(-257\right)=-2020-\left(-1863\right)-\left(-257\right)\)

\(=100\)

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: 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}

20010

cảm ơn bạn nhưng bạn ghi rõ được cho mk ko bạn??
 

B = \(\frac{2001}{2002}+\frac{2002}{2003}\)

có: \(\frac{2000}{2001}>\frac{2000}{2001}+2002\)

\(\frac{2001}{2002}>\frac{2001}{2001}+2002\)

Vậy A>B

minh sinh nam 2005

lam NY ban dc ko

minh 2006

mk sinh năm 2005.

mà mấy bạn cho mk hỏi NY là cái gì?

-388593.198

giải chi tiết đc 0 bạn

Bài 1:

a) \(\left|3x-5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x=-2004\)( do \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\))

Bài 2:

a) \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)

\(=\dfrac{1}{4}+\dfrac{3}{4}=1\)

b) \(=-\left(\dfrac{1}{99.100}+\dfrac{1}{98.99}+\dfrac{1}{97.98}+...+\dfrac{1}{2.3}+\dfrac{1}{1.2}\right)\)

\(=-\left(\dfrac{1}{99}-\dfrac{1}{100}+\dfrac{1}{98}-\dfrac{1}{99}+...+1-\dfrac{1}{2}\right)\)

\(=-\left(1-\dfrac{1}{100}\right)=-\dfrac{99}{100}\)

Bài 1:

a) \(\left|3x-5\right|=4\)  (1)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=4\\3x-5=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=9\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{3}\end{matrix}\right.\)

b) \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)

\(\Leftrightarrow x+1=0\)    \(\left(do\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\right)\)

\(\Leftrightarrow x=-1\)

c) \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)

\(\Leftrightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)           \(\left(do\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

\(\Leftrightarrow x=-2004\)