\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\left(dkxd:x,y\ne0\right)\)

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

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{x^3+y^3}{x^2y^2}\)

`#3107.101107`

`x^3 - 3x^2 - 4x + 12 = 0`

`\Leftrightarrow (x^3 - 3x^2) - (4x - 12) = 0`

`\Leftrightarrow x^2(x - 3) - 4(x - 3) = 0`

`\Leftrightarrow (x^2 - 4)(x - 3) = 0`

`\Leftrightarrow (x - 2)(x + 2)(x - 3) = 0`

`\Leftrightarrow`\(\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\)

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

Vậy, `x \in {-2; 2; 3}`

_______

Sử dụng HĐT: \(A^2-B^2=\left(A-B\right)\left(A+B\right).\)

`#3107.101107`

`(x - 1)(x + 2) - 2x - 4 = 0`

`\Leftrightarrow (x - 1)(x + 2) - (2x + 4) = 0`

`\Leftrightarrow (x - 1)(x + 2) - 2(x + 2) = 0`

`\Leftrightarrow (x + 2)(x - 1 - 2) = 0`

`\Leftrightarrow (x + 2)(x - 3) = 0`

`\Leftrightarrow`\(\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\)

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

Vậy, `x \in {-2; 3}.`

\(\left(x-1\right)\left(x+2\right)-2x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)-\left(2x+4\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0+3\\x=0-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy x=3 hoặc x=-2

Bài 1:

a. ĐKXĐ: $1-x\neq 0; 1+x\neq 0; 1-x^2\neq 0$

$\Leftrightarrow x\neq \pm 1$

b.

\(A=\frac{1+x}{(1-x)(1+x)}+\frac{2(1-x)}{(x+1)(1-x)}-\frac{5-x}{(1-x)(1+x)}\\ =\frac{1+x+(2-2x)-(5-x)}{(1-x)(1+x)}=\frac{-2}{(1-x)(1+x)}=\frac{-2}{1-x^2}\)

Lời giải:

$\frac{A}{B}=\frac{3}{5}\Rightarrow A=\frac{3}{5}B$

$\frac{B}{C}=\frac{7}{11}\Rightarrow C=\frac{11}{7}B$

$\frac{C}{D}=\frac{2}{3}\Rightarrow D=\frac{3}{2}C=\frac{3}{2}.\frac{11}{7}B=\frac{33}{14}B$

$A+B+C+D=1161$

$\frac{3}{5}B+B+\frac{11}{7}B+\frac{33}{14}B=1161$

$B.(\frac{3}{5}+1+\frac{11}{7}+\frac{33}{14})=1161$

$B.\frac{387}{70}=1161$

$B=210$

1008³ - 3.1008².8 + 3.1008.8² - 2⁹

= 1008³ - 3.1008².8 + 3.1008.8² - 8³

= (1008 - 8)³

= 1000³

= 1000000000

giúp mình

Tld

Lời giải:

a. $(3x+1)^2+(2-3x)(2+3x)$

$=(9x^2+6x+1)+(4-9x^2)=6x+5$

b.

$(2a-1)(4a^2+2a+1)-4a(2a^2-3)$

$=(2a)^3-1-(8a^3-12a)=8a^3-1-8a^3+12a=12a-1$

Vừa nãy làm sai:

Làm lại:

Số tiền giảm giá là:

\(12000000\text{x}10\%=1200000\left(\text{đồng}\right)\)

Giá tiền tivi sau khi giảm giá là:

\(12000000-1200000=10800000\left(\text{đồng}\right)\)

Đáp số: \(10800000\left(\text{đồng}\right)\)

                                             Bài giải:

Giá tivi sau khi giảm giá là:

\(\dfrac{12000000\text{x}10}{100}=1200000\left(\text{đồng}\right)\)

Đáp số: \(1200000\text{đồng}\)