Tính:
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)
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29 tháng 4 2021
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
29 tháng 4 2021
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
A
11 tháng 9 2020
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
A
11 tháng 9 2020
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
TD
3 tháng 2 2022
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
TD
3 tháng 2 2022
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>x=\(\dfrac{1}{5}\)
16 tháng 12 2022
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
2
16 tháng 12 2022
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
5 tháng 3 2022
a: \(\Leftrightarrow2x^2-2-3>-5x+\left(2x+1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2-5>-5x+2x^2-6x+x-3\)
\(\Leftrightarrow2x^2-5>2x^2-10x-3\)
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: \(\Leftrightarrow x^2-6x+9+8-4x>x+7\)
\(\Leftrightarrow x^2-10x+17-x-7>0\)
\(\Leftrightarrow x^2-11x+10>0\)
=>x>10 hoặc x<1
PT
5 tháng 3 2022
a: ⇔2x2−2−3>−5x+(2x+1)(x−3)⇔2x2−2−3>−5x+(2x+1)(x−3)
⇔2x2−5>−5x+2x2−6x+x−3⇔2x2−5>−5x+2x2−6x+x−3
⇔2x2−5>2x2−10x−3⇔2x2−5>2x2−10x−3
=>-5>-10x-3
=>5<10x+3
=>10x+3>5
=>10x>2
hay x>1/5
b: ⇔x2−6x+9+8−4x>x+7⇔x2−6x+9+8−4x>x+7
⇔x2−10x+17−x−7>0⇔x2−10x+17−x−7>0
⇔x2−11x+10>0⇔x2−11x+10>0
=>x>10 hoặc x<1
25 tháng 2 2023
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
HT
22 tháng 7 2017
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
HT
22 tháng 7 2017
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
HT
6 tháng 3 2021
a/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\left(2x\right)^2.\left(4x\right)^3}{x^4}}{\dfrac{\left(3x\right)^2\left(5x^2\right)}{x^4}}=\lim\limits_{x\rightarrow\pm\infty}\dfrac{4^4.x}{45}=\pm\infty\)
b/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\sqrt[3]{\dfrac{x^3}{x^3}+\dfrac{2x^2}{x^3}+\dfrac{x}{x^3}}}{\dfrac{2x}{x}-\dfrac{2}{x}}=\dfrac{1}{2}\)
c/ \(=\lim\limits_{x\rightarrow\pm\infty}\dfrac{\dfrac{\sqrt[3]{\left(x^3+2x^2\right)^2}}{x^2}+\dfrac{x\sqrt[3]{x^3+2x^2}}{x^2}+\dfrac{x^2}{x^2}}{\dfrac{3x^2}{x^2}-\dfrac{2x}{x^2}}=\dfrac{1+1+1}{3}=1\)
d/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(-3x\right)^3x^2}{x^5}}{-\dfrac{4x^5}{x^5}}=\dfrac{-27}{-4}=\dfrac{27}{4}\)
e/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{\left(2x\right)^{20}.\left(3x\right)^{20}}{x^{50}}}{\dfrac{\left(2x\right)^{50}}{x^{50}}}=0\)
g/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{8x^3.\left(4x^5\right)^9}{x^{47}}}{\dfrac{11x^{47}}{x^{47}}}=+\infty\)
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)
\(=x^3-2^3-\left(x^3-3x^2+3x-1\right)+7\)
\(=x^3-8-x^3+3x^2-3x+1+7\)
\(=3x^2-3x\)
\(=3x\left(x-1\right)\)
Làm
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)
\(=\left(x^3-2^3\right)-\left(x-1\right)^3+7\)
Vậy...........