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14 tháng 10 2021
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
14 tháng 10 2021
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
17 tháng 9 2021
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
18 tháng 9 2021
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
28 tháng 7 2021
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
28 tháng 7 2021
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
28 tháng 2 2022
\(\Leftrightarrow2x^3-2x+x^2-1-4x^2+2x+2=0\)
\(\Leftrightarrow2x^3-3x^2+1=0\)
\(\Leftrightarrow2x^3-2x^2-x^2+1=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-2x+x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(2x+1\right)=0\)
=>x=1 hoặc x=-1/2
28 tháng 2 2022
\(\left(2x+1\right)\left(x^2-1\right)=4x^2-2x-2\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=4x^2-4x+2x-2\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=4x\left(x-1\right)+2\left(x-1\right)\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=\left(4x+2\right)\left(x-1\right)\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)=2\left(2x+1\right)\left(x-1\right)\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1\right)-2\left(2x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)\left(x+1-2\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
C
1
7 tháng 7 2021
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)
10 tháng 7 2021
a) Ta có: \(\left(x-1\right)\left(x-2\right)\left(x^2+x+1\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x-2\right)\left(x^2+2x+4\right)-x^6+9x^3\)
\(=\left(x^3-1\right)\left(x^3-8\right)-x^6+9x^3\)
\(=x^6-9x^3+8-x^6+9x^3=8\)
b) Ta có: \(\left(\dfrac{1}{3}+2x\right)\left(\dfrac{1}{9}-\dfrac{2}{3}x+4x^2\right)-\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2}{3}x+\dfrac{1}{4}\right)\)
\(=\dfrac{1}{27}+8x^3-8x^3+\dfrac{1}{27}\)
\(=\dfrac{2}{27}\)
c) Ta có: \(\left(x-1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(1-x\right)\)
\(=x^3-3x^2+3x-1-x^3+1-3x+3x^2\)
=0
d) Ta có: \(\left(x^2-y^2\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)-x^6+y^6\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)-x^6+y^6\)
\(=x^6-y^6-x^6+y^6=0\)
NL
1
KV
13 tháng 8 2023
a) A = (x - 3)(x² + 3x + 9) - (x³ + 3)
= x³ - 3³ - x³ - 3
= (x³ - x³) + (-27 - 3)
= -30
b) B = (2x + 1)(4x² - 2x + 1) - 8(x + 1/2)(x² - 1/2 x + 1/4)
= (2x)³ + 1³ - 8[x³ + (1/2)³]
= 8x³ + 1 - 8(x³ + 1/8)
= 8x³ + 1 - 8x³ - 1
= (8x³ - 8x³) + (1 - 1)
= 0
TT
3
13 tháng 11 2021
\(a,\Leftrightarrow x\left(2x-7\right)+2\left(2x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{7}{2}\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-9\right)=0\\ \Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-1\right)\left(2x+1\right)-2\left(2x-1\right)^2=0\\ \Leftrightarrow\left(2x-1\right)\left(2x+1-4x+2\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(-2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(2x^3-2x+x^2-1=4x^2-2x-2\)
\(2x^3-2x+x^2-1-4x^2+2x+2=0\)
\(2x^3-3x^2+1=0\)
\(2x^3-2x^2-x^2+1=0\)
\(2x^2.\left(x-1\right)-\left(x^2-1\right)=0\)
\(2x^2.\left(x-1\right)-\left(x-1\right).\left(x+1\right)=0\)
\(\left(x-1\right).\left(2x^2-x-1\right)=0\)
*) \(x-1=0\Rightarrow x=1\)
*)\(2x^2-x-1=0\Rightarrow2x^2-2x+x-1=0\Rightarrow2x.\left(x-1\right)+\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(2x+1\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}\)
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