\(\left(x-9\right)^3.\left(x^2-4\right)-x^2=-4\)

Đầu bài thế này hả bạn??

đúng rồi đó

\(\left(x-9\right)^3.\left(x^2-4\right)-x^2=-4\)

\(\Leftrightarrow\left(x-9\right)^3.\left(x^2-4\right)-\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[\left(x-9\right)^3-1\right].\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left[\left(x-9\right)^2+x-9+1\right]\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left[x^2-17x+73\right]\left(x+2\right)\left(x-2\right)=0\)

\(\Leftrightarrow x\in\left\{10;2;-2\right\}\)

\(1,\\ a,\Rightarrow x=\dfrac{5\cdot4}{3}=\dfrac{20}{3}\\ b,\Rightarrow x=8\\ c,\Rightarrow x=\dfrac{27\cdot2}{3,6}=15\\ 2,\\ a,\dfrac{6}{9}=\dfrac{42}{63};\dfrac{6}{42}=\dfrac{9}{63};\dfrac{63}{9}=\dfrac{42}{6};\dfrac{63}{42}=\dfrac{9}{6}\\ b,\dfrac{4}{18}=\dfrac{2}{9};\dfrac{4}{2}=\dfrac{18}{9};\dfrac{9}{18}=\dfrac{2}{4};\dfrac{9}{2}=\dfrac{18}{4}\)

a) \(\left(x^2-9\right)\cdot\left(4^x-16\right)=0\)

\(\Rightarrow x^2-9=0\)hoặc \(4^x-16=0\)

               \(x^2=9\)                     \(4^x=16\)

               \(x^2=\left(\pm3\right)^2\)         \(4^x=4^2\)                

\(\Rightarrow x=\pm3\)hoặc \(x=2\)

b) \(5^x+5^{x+2}=650\)

\(\Rightarrow5^x+5^x\cdot25=650\)

\(\Rightarrow5^x\cdot\left(1+25\right)=650\)

\(\Rightarrow5^x\cdot26=650\)

\(\Rightarrow5^x=650\div26=25\)

\(\Rightarrow5^x=5^2\)

\(\Rightarrow x=2\)Vậy \(x=2\)

c) \(2^{x+2}-2^x=96\)

\(2^x\cdot4-2^x=96\)

\(2^x\cdot\left(4-1\right)=96\)

\(2^x\cdot3=96\)

\(2^x=96\div3=32\)

\(2^x=2^5\)Vậy \(x=5\)

a.    12,5 - \(x\) = \(x\) + 12

 \(x\) + 12  + \(x\) = 12,5

  \(x\) x 1 + \(x\) x 1 = 12,5 - 12

  \(x\) x (1 + 1) = 0,5

  \(x\) x 2         = 0,5

  \(x\)               = 0,5 : 2

  \(x\)              = 0,25

b, 8 \(\times\) \(x\) + 3 \(\times\) \(x\) + 2 \(\times\) \(x\) = 20

   \(x\) \(\times\) (8 + 3 + 2)  =20

   \(x\) \(\times\) 13 = 20

  \(x\)          = 20 : 13

  \(x\)          = \(\dfrac{20}{13}\)

b, 8 $\times$ $x$ + 3 $\times$ $x$ + 2 $\times$ $x$ = 20

   $x$ $\times$ (8 + 3 + 2)  =20

   $x$ $\times$ 13 = 20

  $x$          = 20 : 13

  $x$          = 20131320​
 

A)  7/38 x 9/11 +7/38 x 4/11 -7/38 x 2/11

=7/38.(9/11+4/11-2/11)

=7/38

B) 5/31 x 21/25 + 5/31 x -7/10 - 5/31 x 9/20

=5/31.(21/25-7/10-9/20)

=5/31.(-31/100)

=-1/20

1,25.x-0,75.x=2.25-1

0.5.x=1.25

x=2,5

2.5 bàn nha

`a)4/5:3/x xx2/11=24/165`

`(4/5xx2/11)xx 3/x=8/55`

`8/55xx3/x=8/55`

`3/x=8/55:8/55`

`3/x=1`

`3:x=1`

`x=3:1`

`x=3`

____________________________________________

`b)4/3:x/5=5/13xx4/9`

`4/3:x/5=20/117`

`x/5=4/3:20/117`

`x/5=39/5`

`x=39`

A) 4/5:3/x X 2/11=24/165

\(\dfrac{4}{5}:\dfrac{3}{x}X\dfrac{2}{11}=\dfrac{24}{165}\)

\(\dfrac{3}{x}=\dfrac{24}{165}:\dfrac{4}{5}:\dfrac{2}{11}\)

\(\dfrac{3}{x}=1\)

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

=> x=3

B) \(\dfrac{4}{3}:\dfrac{x}{5}=\dfrac{5}{13}x\dfrac{4}{9}\)

\(\dfrac{4}{3}:\dfrac{x}{5}=\dfrac{20}{117}\)

\(\dfrac{x}{5}=\dfrac{4}{3}:\dfrac{20}{117}\)

\(\dfrac{x}{5}=\dfrac{39}{5}\)

=> x = 39

Bạn chú thích hơi quá lố :) 

Ta có :( 5x - 3y + 4z ) . ( 5x - 3y - 4z ) \(=\left(5x-3y\right)^2-16z^2\)

\(=25x^2-30xy+9y^2-16z^2\)

Mà x^2=y^2 + z^2 nên ( 5x - 3y + 4z ) . ( 5x - 3y - 4z )\(=25x^2-30xy+9y^2-16\left(x^2-y^2\right)\)

\(=9x^2-30xy+25y^2=\left(3x-5y\right)^2\)

Học tốt !

\(\left(5x-3y+4z\right)\left(5x-3y-4z\right)=\left(3x-5y^2\right)\)

\(\Leftrightarrow\left(5x-3y\right)^2-16z^2-\left(3x-5y\right)^2=0\)

\(\Leftrightarrow\left(5x-3y-3x+5y\right)\left(5x-3y+3x-5y\right)-16z^2=0\)

\(\Leftrightarrow16x^2=16y^2+16z^2\)(luôn đúng)

 \(a^2-3ab+2c=0\)

\(\Leftrightarrow\left(x+y\right)^2-3\left(x+y\right)\left(x^2+y^2\right)+2\left(x^3+y^3\right)=0\)

\(\Leftrightarrow x^2+y^2+2xy-3x^3-3xy^2-3yx^2-3y^3+2y^3+2x^3=0\)

\(\Leftrightarrow\left(x+y\right)^2-\left(x+y\right)^3=0\)

Hình như đề hơi sai nếu là a^3 thì ms đúng chứ

mk chỉ góp ý thui tại làm mãi ko ra

hhihihihi

đố ai làm được 😈😈😈😈

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