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

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

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

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

=> \(\hept{\begin{cases}3x=1\\x=2\end{cases}}\)

=>\(\hept{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)

vậy x thuộc (1/3;2)

b. 4x(x-3)-2x+6=0

4x(x-3) -2(x-3)=0

(4x-2)(x-3)

=>*4x-2=0

4x=2

x=1/2

*x-3=0

x=3

vậy x thuộc (1/2;3)

Trả lời:

2CxHyOz + (4x+y-2z)/2 O2 -----> 2x CO2 + yH2O

Học tốt

nhầm đề bạn ơi

a. \(^{77^2}\) + \(23^2\) + 77 . 46

= \(77^2\) + 2. 77 . 23 + \(23^2\)

= \(\left(77+23\right)^2\)

= \(100^2\)

= 10000

b. \(105^2\) - \(5^2\)

= ( 105 - 5)( 105 + 5)

= 100 . 110

= 11000

Bài làm:

xFe2O3 + (3x - 2y ) CO (tº) ----> 2 FexOy + (3x- 2y )CO2

Chúc bạn học giỏi

\(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x\left(x^2+2x+4\right)}\)(điều kiện: \(x\ne\left\{0;2\right\}\)

ĐK: \(x\ne\left\{0;2\right\}\)

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

\(x^4-8x\)\(=\left(x-2\right)\left(x^3+2x^2+4x\right)\) (chỗ này mình làm hơi tắt xíu,bạn tự giải ra chi tiết nha)

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

Từ (1) và (2),ta có: \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)\(=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

\(x^2-9x+1=0\Rightarrow x^2+1=9x\)

\(A=\frac{x^4+x^2+1}{5x^2}=\frac{x^4+2x^2+1-x^2}{5x^2}=\frac{\left(x^2+1\right)^2-x^2}{5x^2}=\frac{\left(x^2-x+1\right)\left(x^2+x+1\right)}{5x^2}\)

\(=\frac{\left(9x-x\right)\left(9x+x\right)}{5x^2}=\frac{80x^2}{5x^2}=16\left(x\ne0\right)\)

\(\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=\left(x^2-13x+30\right)\left(x^2-11x+30\right)-24x^2\)

Đặt \(t=x^2-11x+30\)

\(\Rightarrow\left(x-3\right)\left(x-5\right)\left(x-6\right)\left(x-10\right)-24x^2\)

\(=t.\left(t-2x\right)-24x^2\)

\(=t^2-2xt-24x^2\)

\(=\left(t^2-2xt+x^2\right)-25x^2\)

\(=\left(t-x\right)-\left(5x\right)^2\)

\(=\left(t-6x\right)\left(t+4x\right)\)

\(=\left(x^2-17x+30\right)\left(x^2-7x+30\right)\)

Tham khảo nhé~

\(\frac{\frac{x}{x-y}-\frac{y}{x+y}}{\frac{y}{x-y}+\frac{x}{x+y}}\)

\(=\frac{\frac{x-y}{\left(x-y\right)\left(x+y\right)}}{\frac{y+x}{\left(x-y\right)\left(x+y\right)}}\)

\(=\frac{x-y}{x+y}\)

Tham khảo nhé~