\(8.2^5.16=8.32.16=4096=2^{12}\)

Giúp em với ạ 

uk

Bài 3: Buổi chiều bán được: 75:60%=125(kg)

Cả hai buổi bán được:

75+125=200(kg)=2(tạ)

Bài 4:

Số tiền lãi người đó nhận được là:

\(12000000\cdot0,8\%=96000\left(đồng\right)\)

Số tiền người đó rút ra sau 1 tháng là:

12000000+96000=12096000(đồng)

Bài 1:

a: 67,49:17+32,45:2,5

=3,97+12,98

=16,95

b: (10,54+18,46)x16,5

=29x16,5

=478,5

Bài 2:

a: \(4,5\times\left(x+1,25\right)=4,5\)

=>x+1,25=4,5:4,5=1

=>x=1-1,25=-0,25

b: \(x\times4,25=1,7\times34\)

=>x=1,7x34:4,25=13,6

c: \(\left(x+0,75\right)\times\left(4-0,8\right)=205,8\)

=>\(\left(x+0,75\right)\times3,2=205,8\)

=>x+0,75=205,8:3,2=64,3125

=>x=64,3125-0,75=63,5625

Bài 6:

a) Ta có:

\(\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>2\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>A=2\sqrt{x+7}-5\ge-5\forall\left(x\ge-7\right)\)

Dấu "=" xảy ra: `x+7=0`

`<=>x=-7` 

b) Ta có:

\(\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>A=-12+\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\)

Dấu "=" xảy ra: `x-8<=>x=8` 

Bài 7:

a: ĐKXĐ: x>=0

\(-\dfrac{1}{4}\sqrt{x}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x}+4< =4\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

b: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-2\end{matrix}\right.\)

\(\sqrt{x^2-4}>=0\forall x\) thỏa mãn DKXĐ
=>\(-\dfrac{1}{4}\sqrt{x^2-4}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x^2-4}+1< =1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(x^2-4=0\)

=>\(x^2=4\)

=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

g) (3x + 4)(3x - 4) - (2x + 5)² = (x - 5)² + (2x + 1)² - (x² - 2x) + (x - 1)²

9x² - 16 - 4x² - 20x - 25 = x² - 10x + 25 + 4x² + 4x + 1 - x² + 2x + x² - 2x + 1

5x² - 20x - 41 = 5x² - 6x + 27

5x² - 5x² - 20x + 6x = 27 + 41

-14x = 68

i) -5(x + 3)² + (x - 1)(x + 1) + (2x - 3)² = (5x - 2)² - 5x(5x + 3)

-5(x² + 6x + 9) + x² - 1 + 4x² - 12x + 9 = 25x² - 20x + 4 - 25x² - 15x

-5x² - 30x - 45 + x² - 1 + 4x² - 12x + 9 = -35x + 4

-42x - 37 = -35x + 4

-42x + 35x = 4 + 37

-7x = 41

a) (-x + 5)(x - 2) + (x - 7)(x + 7) = (3x + 1)² - (3x - 2)(3x + 2)

-x² + 2x + 5x - 10 + x² - 49 = 9x² + 6x + 1 - 9x² + 4

7x - 59 = 6x + 5

7x - 6x = 5 + 59

x = 64

b) (5x - 1)(x + 1) - 2(x - 3)² = (x + 2)(3x - 1) - (x + 4)² + (x² - x)

5x² + 5x - x - 1 - 2(x² - 6x + 9) = 3x² - x + 6x - 2 - x² - 8x - 16 + x² - x

5x² + 4x - 1 - 2x² + 12x - 18 = 3x² - 4x - 18

3x² + 16x - 19 = 3x² - 4x - 18

3x² + 16x - 3x² + 4x = -18 + 19

20x = 1

Bài 2: 

\(a,x-\dfrac{3}{10}=\dfrac{7}{15}\cdot\dfrac{3}{5}\\ =>x-\dfrac{3}{10}=\dfrac{7}{25}\\ =>x=\dfrac{7}{25}+\dfrac{3}{10}\\ =>x=\dfrac{29}{50}\\ b.2x+\dfrac{3}{2}=\dfrac{-2}{5}\\ =>2x=\dfrac{-2}{5}-\dfrac{3}{2}\\ =>x=\dfrac{-4}{10}-\dfrac{15}{10}=\dfrac{-19}{10}\\ =>x=\dfrac{-19}{10}:2=-\dfrac{19}{20}\\ c,\left(x-\dfrac{1}{2}\right)^3=-8\\ =>\left(x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\\ =>x-\dfrac{1}{2}=-2\\ =>x=-2+\dfrac{1}{2}\\ =>x=\dfrac{-3}{2}\\ d,\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\\ =>\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\\=>x=2\)

Câu 3:

a: \(\widehat{xOz}+\widehat{yOz}=180^0\)(hai góc kề bù)

=>\(\widehat{yOz}=180^0-\widehat{xOz}=180^0-60^0=120^0\)

b: Ot là phân giác của góc yOz

=>\(\widehat{yOt}=\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=\dfrac{120^0}{2}=60^0\)

Ta có: \(\widehat{xOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}+60^0=180^0\)

=>\(\widehat{xOt}=120^0\)

c: Ta có: \(\widehat{xOm}=\widehat{yOt}\)(hai góc đối đỉnh)

mà \(\widehat{yOt}=60^0\)

nên \(\widehat{xOm}=60^0\)

Ta có: \(\widehat{xOm}=\widehat{xOz}\left(=60^0\right)\)

=>Ox là phân giác của góc mOz

Câu 1:

b: \(\dfrac{11}{2}\cdot4\dfrac{5}{3}-2\dfrac{5}{3}\cdot\dfrac{11}{2}\)

\(=\dfrac{11}{2}\left(4+\dfrac{5}{3}-2-\dfrac{5}{3}\right)\)

\(=\dfrac{11}{2}\cdot2=11\)

d: \(\left(\dfrac{3}{7}\right)^0\cdot1^{15}+\dfrac{7}{9}:\left(\dfrac{2}{3}\right)^2-\dfrac{4}{5}\)

\(=1\cdot1+\dfrac{7}{9}:\dfrac{4}{9}-\dfrac{4}{5}\)

\(=1-\dfrac{4}{5}+\dfrac{7}{4}=\dfrac{1}{5}+\dfrac{7}{4}=\dfrac{4}{20}+\dfrac{35}{20}=\dfrac{39}{20}\)

\(\left[\left(-\dfrac{8}{3}\right)^2\right]^{1010}=\left(-\dfrac{8}{3}\right)^{2020}\)

\(\left[\left(\dfrac{7}{3}\right)^{505}\right]^3=\left(\dfrac{7}{3}\right)^{1515}\)