a. 1/3 + 1/4 - 1/6

= 7/12 - 1/6

= 5/12

b. 2/5 x 5/7 : 3/4

= 2/7 : 3/4

= 8/21

a) \(\dfrac{1}{3}+\dfrac{4}{3}\times\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{4}{6}=\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

b) \(\dfrac{3}{5}\times\dfrac{4}{7}:\dfrac{16}{21}=\dfrac{3}{5}\times\dfrac{4}{7}\times\dfrac{21}{16}=\dfrac{12}{35}\times\dfrac{21}{16}=\dfrac{252}{560}=\dfrac{9}{20}\)

`@`Thay `x=2` vào `A` có:

 `A=3^2-9.2=9-18=-9`

`@` Thay `x=1/3` vào `A` có:

`A=(1/3)^2-9. 1/3=1/9-3=-26/9`

Khi x=2 thì \(A=3\cdot2^2-9\cdot2=12-18=-6\)

Khi x=1/3 thì \(A=3\cdot\dfrac{1}{9}-9\cdot\dfrac{1}{3}=\dfrac{1}{3}-3=-\dfrac{8}{3}\)

\(a,\) Số số hạng là \(\left(40-2\right):2+1=20\left(số\right)\)

Tổng là \(\left(40+2\right)\times20:2=420\)

\(b,\) Số số hạng là \(\left(39-1\right):2+1=20\left(số\right)\)

Tổng là \(\left(39+1\right)\times20:2=400\)

a) 2 + 4 + 6 + 8 + ... + 34 + 36 + 38 + 40

= ( 2 + 42 ) + ( 4 + 38 ) + .... + ( 20 + 22 )

= 42 \(\times\) 10

= 420

b) 1 + 3 + 5 + 7 + ... + 35 + 37 + 39

= ( 1 + 39 ) + ( 3 + 37 ) + ...+ ( 19 + 21 )

= 40 \(\times\) 10

= 400

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: Khi x=-2 thì \(A=3\cdot\left(-2\right)^2+5\cdot\left(-2\right)-1=12-10-1=1\)

b: \(B=6xyz^4=6\cdot3\cdot2\cdot1^4=36\)

a) thay x = -2 và biểu thúc ta đc

\(3.\left(-2\right)^2+5.\left(-2\right)-1=12-10-1=1\)

b)thay x = 3 ; y=2 ,z=1 và biểu thúc ta đc

\(9.3.2.1^4+5.3.2.1^4-8.3.2.1^4=3.2.1^4\left(9+5-8\right)=6.6=36\)

a)=1
b)=36

Giúp vs ạ

\(A=-3\left(3+1\right)+5\left(1-1\right)+8\left(-1+1-2\right)\)

\(A=-28\)

A=9x-3x²-5x²-5x+8x²-8x-16

   =-4x-16

   =-4(x-4)

Thay x=-1 vào A, ta được:

A(-1)=-4(-1-4)=(-4).(-5)=20

Vậy A(-1)=20

Lời giải:

$x=\frac{\sqrt{5}-1}{2}$

$2x=\sqrt{5}-1$

$2x+1=\sqrt{5}\Rightarrow (2x+1)^2=5$

$\Leftrightarrow 4x^2+4x-4=0$

$\Leftrightarrow x^2+x-1=0$

Khi đó:
\((4x^5+4x^4-5x^3+2x-2)^2\)

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

\(=(-x^3+2x-2)^2=[-x(x^2+x+1)+(x^2+x-1)-1]^2\)

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

Ta có:

\(A=x\left(x+y\right)-x\left(y-x\right)=x^2+xy-xy+x^2=2x^2\)

Thay \(x=-3\) vào A, ta có:

\(A=2.\left(-3\right)^2=18\)

Vậy A=18

\(A=x\left(x+y\right)-x\left(y-x\right)=x\left(x+y\right)+x\left(x+y\right)=\left(x+y\right).2x=\left(-3+2\right).2.\left(-3\right)=6\)

Giúp vs ạ

\(log_575+log_53=log_5\left(75.3\right)=log_5225\)

\(4log_{12}2+2log_{12}3=log_{12}16+log_{12}9=log_{12}\left(16.9\right)=log_{12}144=log_{12}12^2=2\)

\(\dfrac{1}{3}log_3\dfrac{9}{7}+log_37^{\dfrac{1}{3}}=\dfrac{1}{3}\left(log_3\dfrac{9}{7}+log_37\right)=\dfrac{1}{3}log_3\left(\dfrac{9}{7}.7\right)=\dfrac{1}{3}log_39=\dfrac{2}{3}\)