tham khảo:

99/100

\(\frac{1}{x-1}+\frac{2}{x-2}+\frac{3}{x-3}=\frac{6}{x-6}\)

ĐKXĐ : x ≠ 1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 6

pt <=> \(\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{3x^2-9x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

<=> \(\frac{6x^2-22x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

=> \(\left(x-6\right)\left(6x^2-22x+18\right)=6\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

(bạn tự khai triển rút gọn nhé)

<=> \(6x^3-58x^2+150x-108=6x^3-36x^2+66x-36\)

<=>\(6x^3-58x^2+150x-108-6x^3+36x^2-66x+36=0\)

<=> \(-22x^2+84x-72=0\)

<=> \(11x^2-42x+36=0\)

(pt này lên lớp 9 mới học nên mình dừng tại đây)

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

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

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

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

\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow x=3\)

`(x+1)(x+3)=2x^2-2`

`<=>x^2+x+3x+3=2x^2-2`

`<=>x^2-4x-5=0`

`<=>x^2-5x+x-5=0`

`<=>x(x-5)+(x-5)=0`

`<=>(x-5)(x+1)=0`

`<=>` $\left[ \begin{array}{l}x=5\\x=-1\end{array} \right.$

Vậy `S={5,-1}`

Ta có: \(\left(x+1\right)\left(x+3\right)=2x^2-2\)

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

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

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

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

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

\(\Leftrightarrow\left(x+3\right)\left(5-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

Vậy: S={-3;5}

\(\dfrac{2}{30}\)

`(1 xx 2 xx 3 xx 4)/(3 xx 4 xx 5 xx 6)`

`= (1 xx 2)/(5 xx 6)`

`= 2/12`

`= 1/6`

Thôi mình biết làm rồi cảm ơn mn <3

x( x - 1 )( x - 2 )( x - 3 ) + 1 = 0

<=> [ x( x - 3 ) ][ ( x - 1 )( x - 2 ) ] + 1 = 0

<=> ( x² - 3x )( x² - 3x + 2 ) + 1 = 0

<=> ( x² - 3x + 1 - 1 )( x² - 3x + 1 + 1 ) + 1 = 0

<=> ( x² - 3x + 1 )² - 1 + 1 = 0

<=> ( x² - 3x + 1 )² = 0 <=> x² - 3x + 1 = 0

Δ = b² - 4ac = 9 - 4 = 5 > 0 nên pt có hai nghiệm phân biệt \(x_1=\frac{3+\sqrt{5}}{2};x_2=\frac{3-\sqrt{5}}{2}\)

Vậy S = { \(\frac{3\pm\sqrt{5}}{2}\)}

Dùng kiến thức lớp 9 làm gì hả Quỳnh? Đây là lớp 8 mà.

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

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

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

Đặt \(x^2-3x+1=a\), phương trình trở thành:

\(\left(a-1\right)\left(a+1\right)+1=0\).

\(\Leftrightarrow a^2-1+1=0\).

\(\Leftrightarrow a^2=0\).

\(\Leftrightarrow a=0\).

\(\Leftrightarrow x^2-3x+1=0\).

\(\Leftrightarrow\left(x^2-2.\frac{3}{2}.x+\frac{9}{4}\right)-\frac{5}{4}=0\).

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{5}{4}\).

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{5}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{5}}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{5}}{2}\\x=\frac{3-\sqrt{5}}{2}\end{cases}}\).

Vậy phương trình có tập nghiệm: \(S=\left\{\frac{3\pm\sqrt{5}}{2}\right\}\).

Trả lời

X:1:2.3=60

X:1:2   =60:3

X:1       =20

X:1       =40

X           =40.

\(x:1:2\times3=60\)

\(\Leftrightarrow x:2=60:3\)

\(\Leftrightarrow\frac{x}{2}=20\Rightarrow x=20\times2=40\)