a,3535/3737-x=2121/7474

35/37-x=21/74

x=35/37-21/74

x=49/74

b,x/1968-217/984=187/492

x/1968=187/492+217/984

x/1968=197/328

x=197.1967/328

x=????

\(\frac{3535}{3737}-x=\frac{2121}{7474}\)

                 \(x=\frac{3535}{3737}-\frac{2121}{7474}\)

                 \(x=\frac{7070}{7474}-\frac{2121}{7474}\)

a)3535/3737-x=2121/7474

x=3535/3737-2121/7474

x=49/74

b)x:1968-217/984=187/492

x:1968=187/492+217/984

x:1968=197/328

x=197/328x1968

x=1182

tick nha bạn

`1a)A=(5sqrt5+2sqrt{45}+sqrt5).sqrt5`

`=(5sqrt{5}+2sqrt{9.5}+sqrt5).sqrt5`

`=(5sqrt5+6sqrt5+sqrt5).sqrt5`

`=12sqrt5*sqrt5=60`

`b)B=2sqrt2+1/(2sqrt2)-4sqrt{50}`

`=2sqrt2+1/(2sqrt2)-4.sqrt{25.2}`

`=2sqrt2+1/(2sqrt2)-20sqrt2`

`=(8+1-80)/(2sqrt2)`

`=(-71)/(2sqrt2)`

`=(-71sqrt2)/4`

`c)=1/(1-sqrt2)+1/(1+sqrt2)`

`=(sqrt2+1)/(1-2)+(sqrt2-1)/(2-1)`

`=-sqrt2-1+sqrt2-1=-2`

`2)sqrt{(8-4x)^2}=2`

`<=>|8-4x|=2`

`<=>|4-2x|=1`

`<=>` \(\left[ \begin{array}{l}4-2x=1\\4-2x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=3\\2x=5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=\dfrac52\end{array} \right.\) 

Vậy `S={3/2,5/2}`

\(A=\left(5\sqrt{5}+6\sqrt{5}+\sqrt{5}\right).\sqrt{5}=12\sqrt{5}.\sqrt{5}=12.5=60\)

\(B=2\sqrt{2}+\dfrac{1}{2}\sqrt{2}-10\sqrt{2}=-7,5.\sqrt{2}\)

\(C=\dfrac{1}{1-\sqrt{2}}+\dfrac{1}{1+\sqrt{2}}=\dfrac{1+\sqrt{2}}{-1}+\dfrac{1-\sqrt{2}}{-1}=-1-\sqrt{2}-1+\sqrt{2}=-2\)

Bài 2:

\(\sqrt{\left(8-4x\right)^2}=2\)

*TH1: x < 2

\(\sqrt{\left(8-4x\right)^2}=2\)

\(\Leftrightarrow8-4x=2\Leftrightarrow4x=6\Leftrightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

*TH2: x ≥ 2

\(\sqrt{\left(8-4x\right)^2}=2\)

\(\Leftrightarrow4x-8=2\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)

8: DKXĐ: x-1>=0 và 2-2x>=0

=>x>=1 và x<=1

=>x=1

9: ĐKXĐ: x^2-1>=0 và 4-4x^2>=0

=>x^2>=1 và x^2<=1

=>x^2=1

=>x=1 hoặc x=-1

10: ĐKXĐ: x-1>=0 và 3-x>=0

=>1<=x<=3

B

B

Xét  có  là tia phân giác 

 ( Tính chất đường phân giác trong tam giác)

hay 

Có 

hay 

Vậy 

B

B

f-b-d-c-a-e

\(f\rightarrow b\rightarrow d\rightarrow c\rightarrow a\rightarrow e\)

Thứ tự sắp xếp sẽ là: f-b-d-c-a-e

a. (x + y)² = x² + 2xy + y²

b. (x - 2y)² = x² - 4xy - 4x²

c. (xy² + 1)(xy² - 1) = x²y⁴ - 1

d. (x + y)²(x - y)² = (x² + 2xy + y²)(x² - 2xy + y²) = x⁴ - (2xy + y²)² = x⁴ - (4x²y² + y⁴) = x⁴ - 4x²y² - y⁴

^{Chucs hocj toots}

Câu 2: 

a: \(x^2-4x+4=\left(x-2\right)^2\)

b: \(x^2+10x+25=\left(x+5\right)^2\)

d: \(9\left(x+1\right)^2-6\left(x+1\right)+1=\left(3x+2\right)^2\)

e: \(\left(x-2y\right)^2-8\left(x-2xy\right)+16x^2=\left(x-2y+4x\right)^2=\left(5x-2y\right)^2\)

Câu 7:

a: Ta có: \(A=x^2-2x+7\)

\(=x^2-2x+1+6\)

\(=\left(x-1\right)^2+6\ge6\forall x\)

Dấu '=' xảy ra khi x=1

b: Ta có: \(B=5x^2-20x\)

\(=5\left(x^2-4x+4-4\right)\)

\(=5\left(x-2\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=2