Đáp án B

a) x = 15

b) x = 11

c) x = 68

d) x = 310

(x4)3=x6x19

a) x = 15.              b)      x = 11.        c) x = 68.              d) x = 310

Tìm xx biết:    \left(x^{4}\right)^{3}=\dfrac{x^{19}}{x^{6}}(x4)3=x6x19​

Trả lời: x=x= 

a: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}}:\dfrac{\sqrt{x}-1+1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{5}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}=\dfrac{x-1}{\sqrt{x}}+\dfrac{5}{\sqrt{x}}=\dfrac{x+4}{\sqrt{x}}\)

b: Để A=5 thì \(x+4=5\sqrt{x}\)

=>x=1(loại) hoặc x=16(nhận)

\(\left(a\right):2x-7\sqrt{x}+3=0\left(x\ge0\right)\\ < =>\left(2x-6\sqrt{x}\right)-\left(\sqrt{x}-3\right)=0\\ < =>2\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}-3\right)=0\\ < =>\left(2\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ =>\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{4}\left(TM\right)\\x=9\left(TM\right)\end{matrix}\right.\)

\(\left(b\right):3\sqrt{x}+5< 6\\ < =>3\sqrt{x}< 1\\ < =>\sqrt{x}< \dfrac{1}{3}\\ < =>0\le x< \dfrac{1}{9}\)

\(\left(c\right):x-3\sqrt{x}-10< 0\\ < =>\left(x-5\sqrt{x}\right)+\left(2\sqrt{x}-10\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}-5\right)+2\left(\sqrt{x}-5\right)< 0\\ < =>\left(\sqrt{x}-5\right)\left(\sqrt{x}+2\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}-5< 0\\\sqrt{x}+2>0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}0\le x< 25\\x\ge0\end{matrix}\right.< =>0\le x< 25\)

\(\left(d\right):x-5\sqrt{x}+6=0\left(x\ge0\right)\\ < =>\left(x-2\sqrt{x}\right)-\left(3\sqrt{x}-6\right)=0\\ < =>\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=0\\ < =>\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-3=0\\\sqrt{x}-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=9\\x=4\end{matrix}\right.\left(TM\right)\)

\(\left(e\right):x+5\sqrt{x}-14< 0\\ < =>\left(x+7\sqrt{x}\right)-\left(2\sqrt{x}+14\right)< 0\\ < =>\sqrt{x}\left(\sqrt{x}+7\right)-2\left(\sqrt{x}+7\right)< 0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}+7\right)< 0\\ =>\left\{{}\begin{matrix}\sqrt{x}+7>0\\\sqrt{x}-2< 0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x\ge0\\0\le x< 4\end{matrix}\right.< =>0\le x< 4\)

A.\(\left(x-15\right).15=0\)

\(x-15=0:15\)

\(x-15=0\)

\(x=15+0\)

\(x=15\)

B.\(32\left(x-10\right)=32\)

\(x-10=32:32\)

\(x-10=1\)

\(x=10+1\)

\(x=11\)

`a) `

`(x-15)xx15=0`

`<=> x-15 = 0 : 15`

`<=> x-15 = 0`

`<=> x = 0 + 15`

`<=> x =15`

`b)`

`32.(x-10)=32`

`<=> x - 10 = 32:32`

`<=>x-10=1`

`<=> x = 1+10`

`<=> x =11`

`c)`

`(x-5).(x-7)=0`

`<=>` \(\left[ \begin{array}{l}x-5 = 0\\x-7=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=5\\x=7\end{array} \right.\) 

`d)`

`(x-35)xx35=35`

`<=> x - 35 = 35:35`

`<=> x - 35 = 1`

`<=> x = 1+35`

`<=> x = 36`

Đáp số:

a) x=45

b) x=31

c) x=2

d) x=94

a, 7\(x\).(2\(x\) + 10) =0

    \(\left[{}\begin{matrix}x=0\\2x+10=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=0\\2x=-10\end{matrix}\right.\)

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

Vậy \(x\in\) {-5; 0}

b, -9\(x\) : (2\(x\) - 10) = 0

    9\(x\)                   = 0 

     \(x\)                    = 0 

c, (4 - \(x\)).(\(x\) + 3)  = 0

    \(\left[{}\begin{matrix}4-x=0\\x+3=0\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)

Vậy \(x\in\) {-3; 4}

d, (\(x\) + 2023).(\(x\) - 2024) = 0

     \(\left[{}\begin{matrix}x+2023=0\\x-2024=0\end{matrix}\right.\)

     \(\left[{}\begin{matrix}x=-2023\\x=2024\end{matrix}\right.\)

Vậy \(x\in\) {-2023; 2024}

a) (x + 5)(x - 4) = 0

x + 5 = 0 hoặc  x - 4=  0

x thuộc {-5 ; 4}

b) (x - 10)(x-  3) = 0

x - 10  = 0 hoặc x - 3 = 0

x thuộc {3;10}

c) (3 - x)(x - 3)  = 0

3 - x = 0 ; x - 3 = 0 

< = . x=  3 (thõa mãn cả 2 ĐK)

d) x(x + 1) = 0

x = 0 hoặc x+  1 = 0

=> x = -1

Vậy x thuộc {-1 ; 0} 

a)(x+5)(x-4)=0

nên x+5=0 hoặc x-4=0

x=0-5                 x=0+4

x=-5                   x=4

b)(x-10)(x-3)=0

nên x-10=0 hoặc x-3=0

x=0+10               x=0+3

x=10                   x=3

c)(3-x)(x-3)=0

nên 3-x=0 hoặc x-3=0

x=3-0               x=0+3

x=3

d)x(x+1)=0

nên x=0 hoặc x+1=0

                     x=0-1

                     x=-1

a/(x + 5).(x - 4) = 0

=> x + 5 = 0 => x = -5

hoặc x - 4 = 0 => x = 4

Vậy x = -5 , x = 4

 b/(x - 10).(x - 3) = 0 

=> x - 10 = 0 => x = 10

hoặc x - 3 = 0 => x = 3

Vậy x = 10, x = 3

c/(3 - x).(x - 3) = 0 

=> (3 - x)² = 0 => 3 - x = 0 => x = 3

Vậy x = 3

d/x.(x + 1) = 0

=> x = 0 

hoặc x + 1 = 0 => x = -1

Vậy x = 0 , x = -1