1.a,=(54+45+1).113

=100.113

=11300

b,=(3/7+8/14)+(4/9+10/18)

=1+1

=2

2.a,=13/10+1/3

=49/30

b,=12/9.(1/12+1/6)

=12/9.1/4

=1/3

c,=3/4.3/2

=9/8

d,=3/2-1/3

=7/6

1:tính bằng cách thuận tiện nhất:

a)54 x 113 + 45 x 113 + 113

= 54 x 113 + 45 x 113 + 113x1

=113 x(54+45+1)

= 113x100

=1300

 b)3/7 + 4/9 + 8/14 + 10/18

=(3/7+8/14)+(4/9+10/18)

=    1           + 1

=2

1

A=101

2

X= 1.25

1 a: 101

   b: 1,25
   c: 2020/2021>2021/2022

E = 1 x 2 + 2 x 3 + 3 x 4 + ... + 2000 x 2001

3 x E = 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 3 + ... + 2000 x 2001 x 3

3 x E = 1 x 2 x 3 + 2 x 3 x (4-1) + 3 x 4 x (5-2) + ... + 2000 x 2001 x (2002-1999)

3 x E = (1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ... + 2000 x 2001 x 2002) - ( 1 x 2 x 3 + 2 x 3 x 4 + 3 x 4 x 5 + ... + 1999 x 2000 x 2001)

3 x E = 2000 x 2001 x 2002

E = 2000 x 667 x 2002

E = 2670668000

chúc bạn học tốt nha

E = 1 .2 + 2.3 + 3.4 + ...+ 2000.2001

=> 3E = 1.2.3+2.3.3+3.4.3+...+2000.2001.3

3E = 1.2.(3-0)+ 2.3.(4-1) + 3.4.(5-2)+...+2000.2001.(2002-1999)

3E = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ...+ 2000.2001.2002 - 1999.2000.2001

3E = 2000.2001.2002

\(E=\frac{2000.2001.2002}{3}=2670668000\)

2:

a: \(=\dfrac{1}{3}\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)=-\dfrac{1}{3}\cdot2=-\dfrac{2}{3}\)

1:

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(=-4-\dfrac{1}{4}=-\dfrac{17}{4}\)

Bài 1:

\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)

\(A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)

\(A=\left(7-6-5\right)-\left(\dfrac{3}{4}+\dfrac{5}{4}-\dfrac{7}{4}\right)+\left(\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\right)\)

\(A=-4-\dfrac{3+5-7}{4}+\dfrac{1+4-5}{3}\)

\(A=-4-\dfrac{1}{4}+\dfrac{0}{3}\)

\(A=-\dfrac{16}{4}-\dfrac{1}{4}+0\)

\(A=\dfrac{-16-1}{4}\)

\(A=-\dfrac{17}{4}\)

Bài 2:

\(\dfrac{1}{3}\cdot-\dfrac{4}{5}+\dfrac{1}{3}\cdot-\dfrac{6}{5}\)

\(=\dfrac{1}{3}\cdot\left(-\dfrac{4}{5}-\dfrac{6}{5}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{-4-6}{5}\)

\(=\dfrac{1}{3}\cdot\dfrac{-10}{5}\)

\(=\dfrac{1}{3}\cdot-2\)

\(=-\dfrac{2}{3}\)

`a)100x^2-20x+1`

`=(10x-1)^2`

Thay `x=1/10`

`=>100x^2-20x+1=(1-1)^2=0`

`b)49x^2-42x+10`

`=49*4/49-42*2/7+10`

`=4-12+10=2`

`c)25x^2+40x+16y^2`

`=(5x+4y)^2=(2+3)^2=25`

`a, 3/4 + 1/2 xx 7/2`

`= 3/4 + 7/4`

`=10/4`

`=5/2`

`b, 6/15 - 1/3 : 5/3`

`= 6/15 - 1/3 xx 3/5`

`= 6/15 - 3/15`

`= 3/15`

`=1/5`

`c, x-4/9 = 3/7 : 9/4`

`=> x-4/9= 3/7 xx 4/9`

`=> x-4/9= 12/63`

`=> x-4/9=4/21`

`=> x= 4/21 +4/9`

`=>x= 40/63`

`d, 7/9 xx 3/5 -1/2=1/5`

`->` sao lại bằng có `x` ko vậy ạ?

`a,`

`3/4+1/2 \times 7/2=3/4+7/4=10/4=5/2`

`b,`

`6/15 - 1/3 \div 5/3=6/15-1/5=1/5`

`c,` Tìm x?

`x-4/9=3/7 \div 9/4`

`x-4/9=4/21`

`x=4/21+4/9`

`x=40/63`

`d, 7/9x \times 3/5-1/2=1/5`

`7/9x \times 3/5=1/5+1/2`

`7/9x \times 3/5=7/10`

`7/9x=7/10 \div 3/5`

`7/9x=7/6`

`x=7/6 \div 7/9=3/2`

Bài 3 : 

Vì \(\left(x-2\right)^2\ge0\forall x\)

Nên :  \(A=\left(x-2\right)^2-4\ge-4\forall x\)

Vậy \(A_{min}=-4\) khi x = 2

B1: lấy máy tính mà tính thôi bạn (nhớ lm theo từng bước)

B2: 

a, \(\left|x-\frac{2}{3}\right|-\frac{1}{2}=\frac{5}{6}\)

\(\left|x-\frac{2}{3}\right|=\frac{4}{3}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=\frac{4}{3}\\x-\frac{2}{3}=\frac{-4}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)

b, \(\frac{\left(-2\right)^x}{512}=-32\Rightarrow\left(-2\right)^x=-16384\Rightarrow x\in\varnothing\)

B3:

Vì \(\left(x-2\right)^2\ge0\Rightarrow A=\left(x-2\right)^2-4\ge-4\)

Dấu "=" xảy ra khi x = 2

Vậy GTNN của A = -4 khi x = 2

a/ \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)

\(=-3x^2+7x-4\)

Thay x = 2 vào A được:

\(=-3.2^2+7.2-4=-2\)

Vậy: Giá trị của A khi x = 2 là -2

==========

b/ \(B=126y^3+\left(x-5y\right)\left(x^2+25y^2+5xy\right)\)

\(=126y^3+x^3-125y^3\)

Thay x = -5 và y = -3 vào B được: 

\(126.\left(-3\right)^3+\left(-5\right)^3-125.\left(-3\right)^3=-152\)

Vậy: Giá trị của B tại x = -5 và y = -3 là -152

==========

c/ \(C=a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(=a^3+b^3-\left(a-b\right)^3\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

Thay a = -4 và b = 4 vào C được:

\(2.4^3+3.\left(-4\right)^2.4-3.\left(-4\right).4^2=512\)

Vậy: Giá trị của C tại a = -4 vào b = 4 là 512

a:Ta có: \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)

\(=-3x^2+7x-4\)

\(=-3\cdot2^2+7\cdot2-4\)

\(=-12-4+14=-2\)

c: Ta có: \(C=a^3+b^3-\left(a-b\right)\left(a^2-2ab+b^2\right)\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

\(=2\cdot4^3+3\cdot\left(-4\right)^2\cdot4-3\cdot\left(-4\right)\cdot4^2\)

\(=128+192+192=512\)