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\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)

`C = (x+4)/(x+1) = (x+1+3)/(x+1) = 1+3/(x+1)`

Để `C in ZZ`

`=> x+1 in Ư(3)=(+-1,+-3)`

`@ x+1  =1 => x =0`

`@ x+1=-1 => x = -2`

`@x+1 =3 => x = 2`

`@x+1 =-3 =>x=-4`

`B = (x-4)/(x+2) = (x+2-6)/(x+2) = 1-6/(x+2)`

Để `B in ZZ`

`=> x+2 in Ư(6) = {+-1,+-2,+-3,+-6)`

`@ x+2 =1 => x = -1`

`@x+2 =-1 => x=-3`

`@ x+2 =2 => x=0`

`@ x+2 =-2 => x=-4`

`@x+2 =3 => x = 1`

`@ x +2 =-3 => x = -5`

`@ x+2 =6 => x=4`

`@x+2 =-6 => x= -8`

a) \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,9\right)\)

\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) \(\sqrt{x}=\sqrt{6+4\sqrt{2}}=\sqrt{\left(2+\sqrt{2}\right)^2}=2+\sqrt{2}\)

\(\Rightarrow Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{2}}{\sqrt{2}-1}=\dfrac{\left(3+\sqrt{2}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(=4\sqrt{2}+5\)

c) \(Q=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=1+\dfrac{4}{\sqrt{x}-3}\)

Để \(Q\in Z\Rightarrow4⋮\sqrt{x}-3\Rightarrow\sqrt{x}-3\in\left\{1;2;4;-1;-2;-4\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{4;5;7;2;1\right\}\Rightarrow x\in\left\{16;25;49;4;1\right\}\)

a) Ta có: \(Q=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)

\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

b) Thay \(x=6+4\sqrt{2}\) vào Q, ta được:

\(Q=\dfrac{2+\sqrt{2}+1}{2+\sqrt{2}-3}=\dfrac{3+\sqrt{3}}{\sqrt{2}-1}=\left(3+\sqrt{3}\right)\left(\sqrt{2}+1\right)\)

a)

\(x+4⋮x\)

\(\Rightarrow\left(x+4\right)-x⋮x\)

\(\Rightarrow4⋮x\)

\(\Rightarrow x\in\left\{1;-1;2;-2;4;-4\right\}\)

b) \(\left(75-x\right)\left(x+32\right)\left(x-58\right)=0\)

\(\Rightarrow75-x=0;x+32=0;x-58=0\)

\(\Rightarrow x=75;x=-32;x=58\)

Ngu

Nhu

`a)` Với `x >= 0,x ne 4` có:

`Q=[2(2-\sqrt{x})+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=[4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=[6-3\sqrt{x}]/[(2+\sqrt{x})(2-\sqrt{x})]`

`Q=3/[2+\sqrt{x}]`

`b)` Với `x >= 0,x ne 4` có:

`Q=6/5<=>3/[2+\sqrt{x}]=6/5`

      `=>12+6\sqrt{x}=15`

   `<=>x=1/4` (t/m)

`c)` Với `x >= 0,x ne 4` có:

`Q in Z<=>3/[2+\sqrt{x}] in ZZ`

   `=>2+\sqrt{x} in Ư_{3}`

  Mà `Ư_{3}={+-1;+-3}`

`@2+\sqrt{x}=1=>\sqrt{x}=-1` (Vô lý)

`@2+\sqrt{x}=-1=>\sqrt{x}=-3` (Vô lý)

`@2+\sqrt{x}=-2=>\sqrt{x}=-4` (Vô lý)

`@2+\sqrt{x}=2=>\sqrt{x}=0<=>x=0` (t/m)

Vậy `x=0`

b. (x-2)(x+15)=0

x-2=0 hoặc x+15=0

x=2 hoặc x=-15

a. (x-2)(x+4)=0

x-2=0 hoặc x+4=0

x=2 hoặc x=-4

g. (x-3)(x-5)<0

\(\begin{cases}x-3>0\\x-5< 0\end{cases}\)=>\(\begin{cases}x>3\\x< 5\end{cases}\)=> 3<x<5 Vậy x= 4

a.

\(\left(x-2\right)\times\left(x+4\right)=0\)

               \(x=2\)

               \(x=-4\)

Vậy x = 2 hoặc x = - 4.

b.

\(\left(x-2\right)\times\left(x+15\right)=0\)

               \(x=2\)

                 \(x=-15\)

Vậy x = 2 hoặc x = - 15.

c.

\(\left(7-x\right)\times\left(x+19\right)=0\)

               \(x=7\)

                 \(x=-19\)

Vậy x = 7 hoặc x = -19.

d.

\(-5< x< 1\)

\(x\in\left\{-4;-3;-2;-1;0\right\}\)

e.

\(\left|x\right|< 3\)

\(\left|x\right|\in\left\{0;1;2\right\}\)

\(x\in\left\{-2;-1;0;1;2\right\}\)

Chúc bạn học tốt

e. /x/<3 => -3<x<3

Vậy x=-2;-1;0;1;2

d. -5<x<1

=> x= -4;-3;-2;-1;0

c. (7-x)(x+19)=0

7-x=0 hoặc x+19=0

x=7 hoặc x=-19

Bạn vui lòng viết đề bằng công thức toán để được hỗ trợ tốt hơn!

giúp mình với ạ

a) Ta có: \(P=\dfrac{3x-2\sqrt{x}-4}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\)

\(=\dfrac{3x-2\sqrt{x}-4-x+1-x-4\sqrt{x}-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x-6\sqrt{x}-7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)