a: Ta có: \(2x\left(x-3\right)+x-3=0\)

\(\Leftrightarrow\left(x-3\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(x^2\left(x-6\right)-x^2+36=0\)

\(\Leftrightarrow\left(x-6\right)\left(x^2-x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=3\\x=-2\end{matrix}\right.\)

Câu 1: 

Ta có: \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: S={1;4}

Câu 2: 

Ta có: \(3x^2-7x+3=0\)

\(\Delta=\left(-7\right)^2-4\cdot3\cdot3=49-36=13\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{13}}{6}\\x_2=\dfrac{7+\sqrt{13}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7-\sqrt{13}}{6};\dfrac{7+\sqrt{13}}{6}\right\}\)

Câu 3: 

Ta có: \(5x^2-x-4=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{4}{5}\right\}\)

Câu 4: 

Ta có: \(7x^2+x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{7}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{8}{7}\right\}\)

Câu 1x^2-5x+4=0

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

<=>[x=1;x=4

Câu 2 3x^2-7x+3=0

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

Câu 3 5*x^2 -x-4 = 0

x=-4/5, x=1

Câu 4 7*x^2 +x-8 = 0

x=-8/7, x=1

bn ơi mk giải thế có chỗ nào ko hiểu bn có thể hỏi mk nhé

bạn câu đầu tiền là 0,2 hay 02 vậy ạ

bạn ơi câu đầu tiên là 0 nhé ! mình viết liền quá ! Xin lỗi bạn 

các bạn giúp mình với đi !!!

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\)
\(\Rightarrow-6< x< -4\)
\(\Rightarrow x=-5\)

`(-1)/4 < x/24 < (-1)/6`

`(-6)/24 < x/24 < (-4)/24`

`->x=-5`

a, 

\(x^{15}=x\)\(\Leftrightarrow\hept{\begin{cases}x=-1\\x=1\\x=0\end{cases}}\)

b, \(3+2^{\text{x}-1}=24-\text{[}4^2-\left(2^2-1\right)\text{]}\)

\(\Rightarrow3+2^{\text{x}-1}=24-\left(16-3\right)\)

\(\Rightarrow3+2^{\text{x}-1}=24-13\)

\(\Rightarrow2^{\text{x}-1}=11-3\)

\(\Rightarrow2^{\text{x}-1}=8\)

\(\Rightarrow2^{\text{x}-1}=2^3\)

\(\Rightarrow x-1=3\Rightarrow x=3+1=4\)

Vậy x = 4 

a, Tham khảo : Tìm x biết x^15=x - Lê Nhật Minh

b, \(3+2^{x-1}=24-\left[4^2-(2^2-1)\right]\)

\(\Leftrightarrow3+2^{x-1}=24-\left[16-(4-1)\right]\)

\(\Leftrightarrow3+2^{x-1}=24-\left[16-3\right]\)

\(\Leftrightarrow3+2^{x-1}=24-13\)

\(\Leftrightarrow3+2^{x-1}=11\)

\(\Leftrightarrow2^{x-1}=8\)

\(\Leftrightarrow2^{x-1}=2^3\)

\(\Leftrightarrow x-1=3\Leftrightarrow x=4\)

Vậy x = 4

a) Ta có: x^15 = x

=> \(x\in\left\{0;1\right\}\left(x\in Z\right)\)

KL....

b)

\(3+2^{x-1}=24-\left[4^2-\left(2^2-1\right)\right]\)

\(\Rightarrow2^{x-1}+3=24-\left(16-3\right)\)

\(\Rightarrow2^{x-1}+3=11\)

\(\Rightarrow2^x=8=2^3\)

\(\Rightarrow x=3\left(x\in Z\right)\)

KL...

1) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12=x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2-12\)

\(=x^4+2x^3+4x^2+3x-10=\left(x^4+2x^3\right)+\left(4x^2+8x\right)+\left(-5x-10\right)\)

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

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

2) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left[\left(x+2\right)\left(x+5\right)\right].\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)

Đặt  \(a=x^2+7x+10\) thì ta có :\(a.\left(a+2\right)-24=a^2+2a-24=\left(a^2+2a+1\right)-25=\left(a+1\right)^2-5^2\)

\(=\left(a+1+5\right)\left(a+1-5\right)=\left(a+6\right)\left(a-4\right)\)

Thay a , ta có :

\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)=\left(x^2+7x+16\right).\left(x^2+x+6x+6\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

Bn ơi câu số 2 yêu cầu làm gì vậy

\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)