\(\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\)

\(đk:x+2#0\Leftrightarrow x#\left(-2\right)\)

\(\Leftrightarrow x^2-12=\left(x+2\right)^2\left(3x^2-6x-3\right)\)

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

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

\(\Leftrightarrow x^2-12=3\left(x^4+2x^3-5x^2-12x-4\right)\)

\(\Leftrightarrow x^2-12=3x^4+6x^3-15x^2-36x-12\)

\(\Leftrightarrow3x^4+6x^3-16x^2-36x=0\)

\(\Leftrightarrow x\left(3x^3+6x^2-16x-36\right)=0\)

\(\Leftrightarrow x=0\left(tm\right)\)

đoạn này mắc......

\(2x^2+100x-10000=0\)

\(\Leftrightarrow2x^2+200x-100x-10000=0\)

\(\Leftrightarrow2x\left(x+100\right)-100\left(x+100\right)=0\)

\(\Leftrightarrow\left(2x-100\right)\left(x+100\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x-100=0\\x+100=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=50\\x=-100\end{cases}}\)

\(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\Rightarrow b=\frac{2ac}{a+c}\)

ta có: \(P=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}=\frac{\frac{a^2+3ac}{a+c}}{\frac{2a^2}{a+c}}+\frac{\frac{c^2+3ac}{a+c}}{\frac{2c^2}{a+c}}\)

\(=\frac{a^2+3ac}{2a^2}+\frac{c^2+3ac}{2c^2}=1+\frac{3}{2}\left(\frac{c}{a}+\frac{a}{c}\right)\ge1+\frac{3}{2}\cdot2\sqrt{\frac{c}{a}\cdot\frac{a}{c}}=4\)

Dấu "=" xảy ra khi a=b=c

Đổi 20 phút =1/3h

Gọi vận tốc xe 1 là x( km ) ĐK: x>5

nên vận tốc xe 2 là x-5 (km)

Thời gian mà xe 1 đi từ A đến B là \(\frac{150}{x}\left(h\right)\)

Thời gian mà xe 2 đi từ A đến B là \(\frac{150}{x-5}\left(h\right)\)

Ta có pt sau\(\frac{150}{x-5}-\frac{105}{x}=\frac{1}{3}\)

Làm nốt nhé bạn mình đang vội

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

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

<=> \(\orbr{\begin{cases}3x+2=0\\x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{2}{3}\\x=1\end{cases}}\)

Vậy S = {-2/3; 1}

b) x² - 1 - (x + 5)(2 - x) = 0

<=> x² - 1 - 2x + x² - 10 + 5x = 0

<=> 2x² + 3x - 11 = 0

<=> 2(x² + 3/2x + 9/16 - 97/16) = 0

<=> (x + 3/4)² - 97/16 = 0

<=> \(\orbr{\begin{cases}x+\frac{3}{4}=\frac{\sqrt{97}}{4}\\x+\frac{3}{4}=-\frac{\sqrt{97}}{4}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{\sqrt{97}-3}{4}\\x=-\frac{\sqrt{97}-3}{4}\end{cases}}\)

Vậy S = {\(\frac{\sqrt{97}-3}{4}\); \(-\frac{\sqrt{97}-3}{4}\)

d) x(2x - 3) - 4x + 6 = 0

<=> x(2x - 3) - 2(2x - 3) = 0

<=> (x - 2)(2x - 3) = 0

<=> \(\orbr{\begin{cases}x-2=0\\2x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=\frac{3}{2}\end{cases}}\)

Vậy  S = {2; 3/2}

e)  x³ - 1 = x(x - 1)

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

<=> (x - 1)(x² + x +  1 - x) = 0

<=> (x - 1)(x² + 1) = 0

<=> x - 1 = 0

<=> x = 1

Vậy S = {1}

f) (2x - 5)² - x² - 4x - 4 = 0

<=> (2x - 5)² - (x + 2)² = 0

<=> (2x - 5 - x - 2)(2x - 5 + x + 2) = 0

<=> (x - 7)(3x - 3) = 0

<=> \(\orbr{\begin{cases}x-7=0\\3x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

Vậy S = {7; 1}

h) (x - 2)(x² + 3x - 2) - x³ + 8 = 0

<=> (x - 2)(x² + 3x - 2) - (x- 2)(x² + 2x + 4) = 0

<=> (x - 2)(x² + 3x - 2 - x² - 2x - 4) = 0

<=> (x - 2)(x - 6) = 0

<=> \(\orbr{\begin{cases}x-2=0\\x-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

Vậy S = {2; 6}

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

\(3x.x-3x+2x-2=0\)

\(2x-2=0\)

\(2x=2\)

\(x=1\)

https://diendan.hocmai.vn/threads/sao-minh-hoc-te-he-3-an-wa.231539/

tham khảo nha. mình lười viết