Thực hiện phép nhân và phép chia ở cả 2 vế và so sánh.

Em điền được kết quả như sau:

a: x-2/3=1/15+4/9=3/45+20/45=23/45

=>x=23/45+2/3=23/45+30/45=53/45

b: x-4/6=11/12+3/4=11/12+9/12=20/12=5/3

=>x-2/3=5/3

hay x=7/3

a) \(\left(x-2,5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2,5=3\\x-2,5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5,5\\x=-0,5\end{matrix}\right.\)

b) \(\left(1-x\right)^3=-8=\left(-2\right)^3\)

\(1-x=-2\)

\(x=3\)

c) \(2^4-x=16\)

\(16-x=16\)

\(x=0\)

\(a,\left(x-2,5\right)^2=9\)

\(+,TH1:x-2,5=3\)

\(\Rightarrow x=3+2,5\\\Rightarrow x=5,5\)

\(+,TH2:x-2,5=-3\)

\(\Rightarrow x=-3+2,5\\\Rightarrow x=-0,5\)

\(b,(1-x)^3=-8\)

\(\Rightarrow\left(1-x\right)^3=\left(-2\right)^3\)

\(\Rightarrow1-x=-2\)

\(\Rightarrow x=1-\left(-2\right)\)

\(\Rightarrow x=3\)

\(c,2^4-x=16\)

\(\Rightarrow16-x=16\)

\(\Rightarrow x=0\)

#\(Toru\)

a, \(=x^3\left(x-y\right)+\left(x-y\right)=\left(x^3+1\right)\left(x-y\right)=\left(x+1\right)\left(x^2-x+1\right)\left(x-y\right)\)

b, \(=\left(a^2-2a+1\right)-9b^2=\left(a-1\right)^2-\left(3b\right)^2=\left(a-3b-1\right)\left(a+3b-1\right)\)

\(a,\left(2x+1\right)^2-4\left(x+2\right)^2=9\\ \Leftrightarrow4x^2+4x+1-4\left(x^2+4x+4\right)-9=0\\ \Leftrightarrow4x^2-4x^2+4x-16x+1-16-9=0\\ \Leftrightarrow-12x=24\\ \Leftrightarrow x=\dfrac{24}{-12}=-2\\ b,\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2-x^2+6x-4x=1-9-32\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\\ c,3\left(x+2\right)^2+\left(2x-1\right)^2-7\left(x+3\right)\left(x-3\right)=36\\ \Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-4x+1\right)-7\left(x^2-9\right)=36\\ \Leftrightarrow3x^2+12x+12+4x^2-4x+1-7x^2+63=36\\ \Leftrightarrow3x^2+4x^2-7x^2+12x-4x=36-12-1-63\\ \Leftrightarrow8x=-40\\ \Leftrightarrow x=\dfrac{-40}{8}=-5\)

a: \(2x\left(x-1\right)-x\left(2x-5\right)=9\)

=>\(2x^2-2x-2x^2+5x=9\)

=>3x=9

=>\(x=\dfrac{9}{3}=3\)

b: \(\left(3x-2\right)^2-5\left(x-1\right)\left(x+2\right)=\left(2x-3\right)^2\)

=>\(9x^2-12x+4-5\left(x^2+x-2\right)=4x^2-12x+9\)

=>\(9x^2-12x+4-5x^2-5x+10=4x^2-12x+9\)

=>\(4x^2-17x+14=4x^2-12x+9\)

=>\(-17x+14=-12x+9\)

=>\(-5x=-5\)

=>x=1

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

\(a,\dfrac{1}{3}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{1}{3}+\dfrac{6}{36}=\dfrac{1}{3}+\dfrac{1}{6}=\dfrac{2}{6}+\dfrac{1}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)

\(b,2-\left(\dfrac{2}{5}+\dfrac{2}{3}\right)=2-\left(\dfrac{6}{15}+\dfrac{10}{15}\right)=2-\dfrac{16}{15}=\dfrac{30}{15}-\dfrac{16}{15}=\dfrac{14}{15}\)

\(c,\dfrac{2}{5}:\dfrac{4}{7}+\dfrac{1}{3}=\dfrac{2}{5}\times\dfrac{7}{4}+\dfrac{1}{3}=\dfrac{14}{20}+\dfrac{1}{3}=\dfrac{7}{10}+\dfrac{1}{3}=\dfrac{21}{30}+\dfrac{10}{30}=\dfrac{31}{30}\)

\(d,\dfrac{9}{13}:\dfrac{3}{26}-\dfrac{8}{13}=\dfrac{9}{13}\times\dfrac{26}{3}-\dfrac{8}{13}=\dfrac{234}{39}-\dfrac{8}{13}=6-\dfrac{8}{13}=\dfrac{78}{13}-\dfrac{8}{13}=\dfrac{70}{13}\)

a:=1/3+6/36

=1/3+1/6

=1/2

b: =2-16/15

=14/15

c: =2/5*7/4+1/3

=14/20+1/3

=7/10+1/3

=21/30+10/30=31/30

d: =9/13*26/3-8/13

=3-8/13=31/13

A) 3x² - x(3x - 5) = 9

3x² - 3x² + 5x = 9

5x = 9

x = 9/5

--------------------

B) 5x² + 9x - 2 = 0

5x² + 10x - x - 2 = 0

(5x² + 10x) - (x + 2) = 0

5x(x + 2) - (x + 2) = 0

(x + 2)(5x - 1) = 0

x + 2 = 0 hoặc 5x - 1 = 0

*) x + 2 = 0

x = -2

*) 5x - 1 = 0

5x = 1

x = 1/5

Vậy x = -2; x = 1/5

---------------------

D) 4(5 - 3x) = 5x - 5

20 - 12x = 5x - 5

-12x - 5x = -5 - 20

-17x = -25

x = 25/17

--------------------

E) 2x² - 11x + 14 = 0

2x² - 4x - 7x + 14 = 0

(2x² - 4x) - (7x - 14) = 0

2x(x - 2) - 7(x - 2) = 0

(x - 2)(2x - 7) = 0

x - 2 = 0 hoặc 2x - 7 = 0

*) x - 2 = 0

x = 2

*) 2x - 7 = 0

2x = 7

x = 7/2

Vậy x = 2; x = 7/2

Câu C và F ghi đề bằng công thức đúng lại em

c: =>x^2-(x-2)(x+5)=2x-1

=>x^2-x^2-5x+2x+10=2x-1

=>3x+10=2x-1

=>x=-11

f: =>3x-5=4(2x+3)

=>8x+12=3x-5

=>5x=-17

=>x=-17/5