\(VT=\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^1-1\right)\left(2^1+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^6+1\right)\left(2^8+1\right)\)

\(=\left(2^{16}-1\right)\left(2^6+1\right)\left(2^8+1\right)\)

tiếp

a;b bất kì, trong đó a; b thuộc N*, a = b

Bài 1:

a) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-3x^2=42\) (1)

\(\Leftrightarrow x^3+9x^2+27x+27-x\left(9x^2+6x+1\right)+8x^3+1-3x^2=42\)

\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=42\)

\(\Leftrightarrow26x+28=42\)

\(\Leftrightarrow26x=42-28\)

\(\Leftrightarrow26x=14\)

\(\Leftrightarrow x=\dfrac{7}{13}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{7}{13}\right\}\)

lấy mt bấm phắt là ra

1) \(C=-9x^2+2x-17\)

\(=-\left(9x^2-2x+17\right)\)

\(=-\left[\left(3x\right)^2-2\cdot x\cdot1+1^2-1+17\right]\)

\(=-\left[\left(3x-1\right)+16\right]\)

\(=-\left(3x-1\right)-16\le-16\)

Vậy biểu thức A luôn âm với mọi giá trị của biến.

2) \(D=-5x^2-6x-11\)

\(=-\left(5x^2+6x+11\right)\)

\(=-\left(5x^2+2\cdot x\cdot3+3^2-9+11\right)\)

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

\(=-\left(x-3\right)^2-2\le2\)

Vậy biểu thức D luôn luôn âm với mọi giá trị của biến.

3) \(E=-\dfrac{1}{4}x^2+3x-15\)

\(=-\left(\dfrac{1}{4}x^2-3x+15\right)\)

\(=-\left[-\left(\dfrac{1}{2}x\right)^2-2\cdot\dfrac{1}{2}x\cdot1+1^2-1+15\right]\)

\(=-\left[-\left(\dfrac{1}{2}x-1\right)+14\right]\)

\(=\left(\dfrac{1}{2}x-1\right)+14\ge14\)

Vậy biểu thức D luôn dương với mọi giá trị của biến.

\(\left|x-2\right|+3=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2+x=-3\left(đk:x-2\ge0\right)\\-\left(x-2\right)+x=3\left(đk:x-2< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\left(đk:x\ge2\right)\\x\in\varnothing\left(đk:x< 2\right)\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)=\frac{-17}{21}:\frac{17}{20}=\frac{-20}{21}\)

\(1-\frac{1}{2}\)\(+\frac{1}{3}-\frac{1}{4}\)=\(\frac{1}{2}+\frac{1}{12}=\frac{7}{12}\)

tới đó rồi chắc bạn làm đc phải ko ???

a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)

\(\Leftrightarrow3-2x=-1\)

\(\Leftrightarrow-2x=-1-3\)

\(\Leftrightarrow-2x=-4\)

\(\Leftrightarrow x=2\)

Vậy \(x=2\)

b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)

\(\Leftrightarrow200x+24=120x-105\)

\(\Leftrightarrow80x=-129\)

\(\Leftrightarrow x=-\dfrac{129}{80}\)

Vậy \(x=-\dfrac{129}{80}\)

c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)

\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)

\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)

\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)

d) \(\left(x+0,2\right)^2+0,75=1\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)

\(\left|x\left(x-4\right)\right|=x\)

\(\Leftrightarrow\left|x^2-4x\right|=x\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x=x\\x^2-4x=-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-x=x-x\\x^2-4x-\left(-x\right)=-x-\left(-x\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=0\\x^2-3x=0\end{matrix}\right.\)

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

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

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

Vậy chọn đáp án A.