a) \(1-\dfrac{1}{1-\dfrac{1}{x}}=1-\dfrac{1}{\dfrac{x-1}{x}}=\dfrac{x-1-x}{x-1}\)

\(=\dfrac{-1}{x-1}\)

b) \(\dfrac{\dfrac{x}{x+1}-\dfrac{x-1}{x}}{\dfrac{x}{x-1}-\dfrac{x-1}{x}}=\dfrac{\dfrac{x^2-\left(x+1\right)\left(x-1\right)}{x\left(x+1\right)}}{\dfrac{x^2-\left(x-1\right)^2}{x\left(x-1\right)}}\)

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

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

Cho x mol Mg và y mol Zn vào dung dịch chứa m mol Cu 2 + và n mol Ag + . Biết rằng  x > 1 2 n . Sau phản ứng thu được dung dịch chứa 3 ion kim loại. Giá trị của y cần thỏa mãn điều kiện nào sau đây ?

A. y > m - x

B. y < m-x

C. y > m – x +  1 2 n

D. y < m – x +  1 2 n

Choose the letter A, B, C or D to complete the sentences

Question: China is famous __________ the Great Wall

A. as

B. for

C. at

D. in

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

\(\Leftrightarrow4x^2-4x+1-2x+1=0\)

\(\Leftrightarrow4x^2-6x+2=0\)

\(\Delta=\left(-6\right)^2-4.4.2=4\)

\(x_1=\dfrac{1}{2};x_2=1\)

b) \(\left(4x-3\right)^2-12x+9=0\)

\(\Leftrightarrow16x^2-24x+9-12x+9=0\)

\(\Leftrightarrow16x^2-36x+18=0\)

\(\Delta=\left(-36\right)^2-4.16.18=144\)

\(x_1=\dfrac{3}{4};x_2=\dfrac{3}{2}\)

c) \(3x^2-48=0\)

\(\Leftrightarrow3x^2=48\)

\(\Leftrightarrow x^2=16\)

\(x_1=4;x_2=-4\)

Tick cho mk nha

\(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)

\(=8x^3+27-8x^3+2\)

\(=29\)

Vậy biểu thức trên ko phụ thuộc biến x

lấy máy tính thay số vào rồi tính thôi

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

b) Nhầm đề nha

\(x^2+xy-x-y=x\left(x+y\right)-\left(x+y\right)=\left(x-1\right)\left(x+y\right)\)

Tick mk nha

\(x+2\sqrt{2}x^2+2x^3=0\)

\(\Leftrightarrow x\left(1+2\sqrt{2}x+2x^2\right)=0\)

\(\Leftrightarrow x\left(1+\sqrt{2}x\right)^2=0\)

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

Vậy x₁ = 0 ; x₂ = \(-\dfrac{\sqrt{2}}{2}\)

3)

a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8\)

\(\Leftrightarrow-2x=7\)

\(\Rightarrow x=\dfrac{-7}{2}\)

b) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2\right)-5x+1=28\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-10x^2+2x+4x^2-5x+1=28\)

\(\Leftrightarrow0-3x^2+23x+28=28\)

\(\Leftrightarrow-3x^2+23x=0\)

\(\Leftrightarrow-x\left(3x-23\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-23=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{3}\end{matrix}\right.\)

c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6-2x^4-2x^2-1=0\)

\(\Leftrightarrow-5x^4+x^2-2=0\)

Đặt \(-5t^2+t-2=0\)

\(\Delta=1^2-4\left(-5\right)\left(-2\right)=-39< 0\)

\(\Rightarrow PTVN\)

2)

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

b) \(\left(x^4-3x^2+9\right)\left(x^2+3\right)-\left(3+x^2\right)=\left(x^2+3\right)\left[x^4-3x^2+9-\left(x^2+3\right)^2\right]\)c) \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=-3x^2+39x+6\)