\(x-7\sqrt{x}+6\)

\(=x-\sqrt{x}-6\sqrt{x}+6\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-6\left(\sqrt{x}-1\right)\)

\(=\left(\sqrt{x}-6\right)\left(\sqrt{x}-1\right)\)

Đề bài là gì vậy bạn ?

\(2x^2-7\sqrt{x}+5\)

\(=2x^2+2x\sqrt{x}+2x-5\sqrt{x}-2x\sqrt{x}-2x-2\sqrt{x}+5\)

\(=\left(2x^2+2x\sqrt{x}+2x-5\sqrt{x}\right)-\left(2x\sqrt{x}+2x+2\sqrt{x}-5\right)\)

\(=\sqrt{x}\left(2\sqrt{x}+2x+\sqrt{x}-5\right)-\left(2x\sqrt{x}+2x+2\sqrt{x}-5\right)\)

\(=\left(\sqrt{x}-1\right)\left(2x\sqrt{x}+2x+2\sqrt{x}-5\right)\)

a: \(\Leftrightarrow4x^2+9x-4x-9=0\)

=>(4x+9)(x-1)=0

=>x=1 hoặc x=-9/4

b: \(\Leftrightarrow x^2-x-4x+4=0\)

=>(x-1)(x-4)=0

=>x=1 hoặc x=4

c: \(\Leftrightarrow5x^2-5x-12x+12=0\)

=>(x-1)(5x-12)=0

=>x=12/5 hoặc x=1

d: \(\Leftrightarrow x^2-4x+x-4=0\)

=>(x-4)(x+1)=0

=>x=4 hoặc x=-1

a, Ta có a + b + c = 4 + 5 - 9 = 0

vậy pt có 2 nghiệm x = 1 ; x = -9/4 

b, Ta có a + b + c = 1 - 5 + 4 = 0 

vậy pt có 2 nghiệm x = 1 ; x = 4 

c, Ta có a + b + c = 5 - 17 + 12 = 0 

vậy pt có 2 nghiệm x = 1 ; x = 12/5 

d, Ta có a - b + c = 1 + 3 - 4 = 0 

vậy pt có 2 nghiệm x = -1 ; x = 4 

\(pt\Leftrightarrow\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-16}{46}-14=0\)

\(\Leftrightarrow\frac{5x-150}{50}-1+\frac{5x-102}{49}-2+\frac{5x-56}{48}-3+\frac{5x-12}{47}-4+\frac{5x-16}{46}-4=0\)

\(\Leftrightarrow\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)

\(\Leftrightarrow\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)

Do \(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\ne0\) nên \(5x-200=0\Rightarrow x=\frac{200}{5}=40\)

Vậy x= 40

\(\frac{5x-150}{50}+\frac{5x-102}{49}+\frac{5x-56}{48}+\frac{5x-12}{47}+\frac{5x-660}{46}=0\)

\(\Leftrightarrow\)\(\left(\frac{5x-150}{50}-1\right)+\left(\frac{5x-102}{49}-2\right)+\left(\frac{5x-56}{48}-3\right)+\left(\frac{5x-12}{47}-4\right)+\left(\frac{5x-660}{46}+10\right)=0\)

\(\Leftrightarrow\)\(\frac{5x-200}{50}+\frac{5x-200}{49}+\frac{5x-200}{48}+\frac{5x-200}{47}+\frac{5x-200}{46}=0\)

\(\Leftrightarrow\)\(\left(5x-200\right)\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}\right)=0\)

\(\Leftrightarrow\)\(5x-200=0\)

\(\Leftrightarrow\)\(5x=200\)

\(\Leftrightarrow\)\(x=40\)

Vậy  x = 40

canh thiu các bạn nhìu(thank you) 

bài này mk hỏi cô nhưng cô chưa trả lời

Ta có : \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)

\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)

\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)

\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)

\(\Leftrightarrow5x-200=0\)

\(\Leftrightarrow x=40\)

Vậy ...

Ta có: \(\dfrac{5x-150}{50}+\dfrac{5x-102}{49}+\dfrac{5x-56}{48}+\dfrac{5x-12}{47}+\dfrac{5x-660}{46}=0\)

\(\Leftrightarrow\dfrac{5x-150}{50}-1+\dfrac{5x-102}{49}-2+\dfrac{5x-56}{48}-3+\dfrac{5x-12}{47}-4+\dfrac{5x-660}{46}+10=0\)

\(\Leftrightarrow\dfrac{5x-200}{50}+\dfrac{5x-200}{49}+\dfrac{5x-200}{48}+\dfrac{5x-200}{47}+\dfrac{5x-200}{46}=0\)

\(\Leftrightarrow\left(5x-200\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}\right)=0\)

mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{46}>0\)

nên 5x-200=0

\(\Leftrightarrow5x=200\)

hay x=40

Vậy: S={40}

a, x : 12 = 27

=> x = 27.12 = 324

Vậy x = 324

b, 1414 : x = 14

=> x = 1414 : 14 = 101

Vậy  x = 101

c, 5x : 12 = 0

=> 5x = 0

=> x = 0.

Vậy x = 0

d, 0 : x = 0

∀ x ∈ ¥, x≠0

a, \(4x\left(x+1\right)-5\left(x+1\right)=0\)

\(\left(x+1\right)\left(4x-5\right)\)=0

\(\left\{{}\begin{matrix}x+1=0\\4x-5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\left(-1\right)\\4x=5\Rightarrow x=\frac{5}{4}\end{matrix}\right.\)

b, \(5x\left(x-20\right)+5x-100=0\)

\(5x\left(x-20\right)+\left(5x-100\right)=0\)

\(5x\left(x-20\right)+5\left(x-20\right)=0\)

\(\left(x-20\right)\left(5x+5\right)\)= 0

\(\left\{{}\begin{matrix}x-20=0\\5x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\5x=-5\Rightarrow x=-1\end{matrix}\right.\)

c, \(2\left(x-2\right)+\left(x-2\right)^2=0\)

Rút gọn thừa số chung

Giải phương trình

Giải phương trình

Biệt thức

Biệt thức

Nghiệm

Lời giải thu được

Vậy x= 0 và x = 2

d, \(\left(x-3\right)^2-5x-x^2=12\)

\(\left(x^2-2.x.3+3^2\right)-5x-x^2=12\)

\(x^2-6x+9-5x-x^2=12\)

\(-11x+9=12\)

\(-11x=3\)

=> \(x=-\frac{3}{11}\)

\(5x:12=0\)

\(\Leftrightarrow5x=0\)

\(\Leftrightarrow x=0\)

`3x^2-5x-12=0`

`<=>3x^2-9x+4x-12=0`

`<=>3x(x-3)+4(x-3)=0`

`<=>(x-3)(3x+4)=0`

`<=>[(x-3=0),(3x+4=0):}<=>[(x=3),(x=-4/3):}`

`3x^2 -5x-12=0`

`<=>3x^2 -9x+4x-12=0`

`<=>3x(x-3)+4(x-3)=0`

`<=>(x-3)(3x+4)=0`

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

Vậy `S={3; -4/3}`

\(3x^2-5x-12=0\)

\(\Leftrightarrow\left(3x+4\right)\left(x-3\right)=0\)

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

Vậy \(S=\left\{-\dfrac{4}{3};3\right\}\).