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:

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

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

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

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

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

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

\(a,\) 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

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

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

Bài 1:

a: \(M=x^2-10x+3\)

\(=x^2-10x+25-22\)

\(=\left(x^2-10x+25\right)-22\)

\(=\left(x-5\right)^2-22>=-22\forall x\)

Dấu '=' xảy ra khi x-5=0

=>x=5

b: \(N=x^2-x+2\)

\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi x-1/2=0

=>x=1/2

c: \(P=3x^2-12x\)

\(=3\left(x^2-4x\right)\)

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

\(=3\left(x-2\right)^2-12>=-12\forall x\)

Dấu '=' xảy ra khi x-2=0

=>x=2

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:
\((4x^5+4x^4-5x^3+2x-2)^2\)

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

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

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

a,= a\(^2\)+2a+b\(^2\)-2b-2ab+37

=a\(^2\)-2ab+b\(^2\)+2a-2b+37

=(a-b)\(^2\)+2(a-b)+37

⇒5\(^2\)+2.5+37= 25+10+37= 72

b,= a\(^3\)+a\(^2\)-b\(^3\)+b\(^2\)+ab-3a\(^2\)b+3ab\(^2\)-3ab-95

=a\(^3\)-3a\(^2\)b+3ab\(^2\)-b\(^3\)+a\(^2\)-2ab+b\(^2\)-95

=(a-b)\(^3\)+(a-b)\(^2\)-95

⇒5\(^3\)+5\(^2\)-95= 125+25-95= 60

`@`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: \(17628+3547\cdot6\)

\(=17628+21282\)

\(=38910\)

b: \(57924-15760:5\)

\(=57924-3152\)

=54772