`1, -2/9 xx 15/17 + (-2/9) xx 2/17`

`= -2/9 xx (15/17 + 2/17)`

`= -2/9 xx 17/17`

`=-2/9xx1`

`=-2/9`

__

`-5/3 xx 6/5 + (-7/9) xx 3/10`

`= -30/15 + (-21/90)`

`= -2 + (-7/30)`

`=-60/30 +(-7/30)`

`=-67/30`

__

`15/20 xx 7/5 + (-9/7) xx (-6/4)`

`=3/4 xx7/5 + (-9/7) xx(-6/4)`

`= 21/20 + 54/28`

`= 21/20 + 27/14`

`=417/140`

__

`-25/13 xx 5/19 + (-25/13) xx 14/19`

`=-25/13 xx (5/19 +14/19)`

`=-25/13 xx 19/19`

`= -25/13 xx 1`

`=-25/13`

__

`-7/13 xx 13/5 + (-9/7) xx 5/3`

`=-7/5 +(-15/7)`

`=-124/35`

1) Rút gọn biểu thức M: M = (2√x)/(√x - 3) - (x + 9√x)/(x - 9) = (2√x(x - 9) - (x + 9√x)(√x - 3))/(√x - 3)(x - 9) = (2x√x - 18√x - x√x + 9x + 9x - 27√x - 9√x + 27 )/(√x - 3)(x - 9) = (2x√x - 36√x + 27x)/(√x - 3)(x - 9) = (x(2√x - 36) + 27x) /(√x - 3)(x - 9) = (x(2√x - 36 + 27))/(√x - 3)(x - 9) = (x(2√x - 9))/( √x - 3)(x - 9) Do đó biểu thức M Rút gọn là: M = (x(2√x - 9))/(√x - 3)(x - 9) 2) Tìm các giá trị của x ă mãn M/N.(căn x + 3) = 3x - 5: Ta có phương trình: M/N.(căn x + 3) = 3x - 5 Đặt căn x + 3 = t, t >= 0, ta có x = t^2 - 3 Thay x = t^2 - 3 vào biểu thức M/N, ta có: M/N = [(t^2 - 3)(2√(t^2 - 3) - 9)]/[(t^2 - 3 + 5)t] = [(2(t^2 - 3) √(t^2 - 3) - 9(t^2 - 3))]/(t^3 + 2t - 3t - 6) = [2(t^2 - 3)√(t^2 - 3) - 9(t^2 - 3)]/(t(t - 1)(t + 2)) Đặt u = t^2 - 3, ta có: M/N = [2u√u - 9u]/((u + 3)(u + 2)) = [u(2√u - 9)]/((u + 3)(u + 2)) Đặt v = √u, ta có: M/N = [(v^ 2 + 3)(2v - 9)]/[((v^2 + 3)^2 - 3)(v^2 + 2)] = [(2v^3 - 18v + 6v - 54)]/[ ( (v^4 + 6v^2 + 9) - 3)(v^2 + 2)] = (2v^3 - 12v - 54)/(v^4 + 6v^2 + 6v^2 - 9v^2 + 18) = (2v^3 - 12v - 54)/(v^4 + 12v^2 + 18) Ta cần tìm các giá trị của v đối xứng phương trình M/N = 3x - 5: (2v^3 - 12v - 54)/(v^4 + 12v^2 + 18) = 3(t^2 - 3) - 5 (2v ^3 - 12v - 54)/(v^4 + 12v^2 + 18) = 3t^ 2 - 14 (2v^3 - 12v - 54) = (v^4 + 12v^2 + 18)(3t^2 - 14) Tuy nhiên, từ t = √(t^2 - 3), ta có v = √u = √(t^2 - 3) => (2(v^2)^3 - 12(v^2) - 54) = ((v^2)^4 + 12(v^2)^2 + 18) (3(v^2 - 3) - 14) => 2v^

\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)

1: =-2/9(15/17+2/17)=-2/9

2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)

=-2-7/30=-67/30

3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

1: =-2/9(15/17+2/17)=-2/9

2: =34⋅75+97⋅32=34⋅75+97⋅32

=21/20+27/14=417/140

4: =-25/13(5/19+14/19)=-25/13

5: =-7/5-45/21=-7/5-15/7=-124/35

Ta có \(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

=> \(\frac{9}{25}x\left(1+\frac{3}{5}+1\right)+18\left(\frac{3}{5}+1\right)=x\)

=> \(\frac{117}{125}x+28,8=x\)

=> \(x-\frac{117}{125}x=28,8\)

=> \(\frac{8}{125}x=28,8\)

=> x = 450

Vậy x = 450

\(\frac{9}{25}x+\frac{3}{5}.\frac{9}{25}x+\frac{3}{5}.18+\frac{9}{25}x+18=x\)

\(x\left(\frac{9}{25}+\frac{9}{25}+\frac{9}{25}\right).\frac{3}{5}+\frac{3}{5}.18+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+18\right)+18=x\)

\(x.\frac{3}{5}\left(\frac{27}{25}+\frac{450}{25}\right)+18=x\)

\(x.\frac{3}{5}.\frac{477}{25}+18=x\)

\(x.\frac{1431}{125}+\frac{2250}{125}=x\)

\(x.\frac{3681}{125}=x\)

vậy chac tui làm sai rồi

Ai biết chỉ mình nha 

1. 11(x-9)=77

x-9=7

x=16

2. 4(x-3)=48

x-3=12

x=15

3.3(x-8)=81

(x-8)=27

x=35

\(40-21:\left(2.x+3\right)=33\)

\(21:\left(2.x+3\right)=40-33\)

\(21:\left(2.x+3\right)=7\)

\(\left(2.x+3\right)=21:7\)

\(\left(2.x+3\right)=3\)

\(2.x=3-3\)

\(2.x=0\)

\(x=0:2\)

\(x=0\)

\(\dfrac{x}{3}=\dfrac{7}{25}-\dfrac{1}{5}\Leftrightarrow\dfrac{25x}{75}=\dfrac{21}{75}-\dfrac{15}{75}\Leftrightarrow25x=21-15\Rightarrow25x=6\Rightarrow x=0,24\)

a) Ta có: \(\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-1}{5}\)

\(\Leftrightarrow\dfrac{x}{3}=\dfrac{7}{25}+\dfrac{-5}{25}=\dfrac{2}{25}\)

hay \(x=\dfrac{6}{25}\)

Vậy: \(x=\dfrac{6}{25}\)

b) Ta có: \(\dfrac{4}{9}+\dfrac{x}{5}=\dfrac{5}{11}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{5}{11}-\dfrac{4}{9}=\dfrac{45}{99}-\dfrac{44}{99}=\dfrac{1}{99}\)

hay \(x=\dfrac{5}{99}\)

Vậy: \(x=\dfrac{5}{99}\)

x/3=7/15+-1/5                             

x/3=2/15                                       

5x/15=2/15                                    

5x=2                                                  

x=2/5

x= 2/5

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