A=(x+1)^2+(x-1)^2-2(x^2-1)
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a) (x+2)(x^2-2x+4)-x(x^2+2)=15
<=> x^3 + 8 - x^3 - 2x = 15
<=> -2x = 7
<=> x = -7/2
b) (x+3)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=28
<=> x^3 + 9x² + 27x + 27 - x(9x² + 6x + 1) + 8x^3 + 1 = 28
<=> x^3 + 9x² + 27x + 27 - 9x^3 - 6x² - x + 8x^3 + 1 - 28 = 0
<=> 3x² + 26x = 0
<=> x(3x + 26) = 0
Vậy x = 0 và x = -26/3
c) (x^2-1)^3-(x^4+x^2+1)(x^2-1)=0
<=> (x² - 1)[(x² -1)² - x^4 - x² - 1] = 0
<=> (x-1)(x+1)(x^4 - 2x² + 1 - x^4 - x² - 1 ) = 0
<=> -(x-1)(x+1)3x² = 0
Vậy nghiệm là x = 1 ; -1 ; 0
a) \(\left|x-\dfrac{1}{2}\right|\le\dfrac{1}{3}\)
\(\Leftrightarrow-\dfrac{1}{3}\le x-\dfrac{1}{2}\le\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{1}{6}\le x\le\dfrac{5}{6}\)
b) \(\left|2x-\dfrac{1}{2}\right|>\left|-1,5\right|\)
\(\Leftrightarrow\left|2x-\dfrac{1}{2}\right|>\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{1}{2}>\dfrac{3}{2}\\2x-\dfrac{1}{2}< \dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x>2\\2x< 1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x>1\\x< \dfrac{1}{2}\end{matrix}\right.\)
=>3x(1-1/6)=3/4
=>3x=3/4:5/6=3/4*6/5=18/20=9/10
=>x=3/10
Đề là \(3x\left(\dfrac{1}{1}\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+\dfrac{1}{4}\times\dfrac{1}{5}+\dfrac{1}{5}\times\dfrac{1}{6}\right)=\dfrac{3}{4}?\)
`3x (1/1 \times 1/2 + 1/2 \times 1/3 + 1/3 \times 1/4 + 1/4 \times 1/5 + 1/5 \times 1/6) = 3/4`
\(3x\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}\right)=\dfrac{3}{4}\)
\(3x\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{3}{4}\)
\(3x\left(1-\dfrac{1}{6}\right)=\dfrac{3}{4}\)
\(3x\times\dfrac{5}{6}=\dfrac{3}{4}\)
`3x=3/4 \div 5/6`
`3x = 9/10`
`x = 9/10 \div 3`
`x = 3/10.`
\(A=x^2+2x+1+x^2-2x+1-2x^2+2=4\)