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1 tháng 11 2018

\(\left(x+1\right)^2+\left(2x-1\right)\left(2x+1\right)-5\left(x-1\right)=77\)

\(\Rightarrow x^2+2x+1+\left(2x\right)^2-1^2-5x+5=77\)

\(\Rightarrow x^2+2x+1+4x^2-1-5x+5=77\)

\(\Rightarrow5x^2-3x+5-77=0\)

\(\Rightarrow5x^2-3x-72=0\)

.....

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

16 tháng 12 2022

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`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`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

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, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)

\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)

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 ~~~ 

11 tháng 11 2021

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

11 tháng 11 2021

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

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: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)

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: \(S=\left\{\dfrac{4}{7}\right\}\)

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

<=>\(x=\dfrac{1}{5}\)

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=\(\dfrac{1}{5}\)

AH
Akai Haruma
Giáo viên
26 tháng 5 2020

Lời giải:

a)

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

\(\Leftrightarrow (x-2)(x-2-1)+2x=(x-2)^2-2\)

\(\Leftrightarrow (x-2)^2-(x-2)+2x=(x-2)^2-2\)

\(\Leftrightarrow x+4=0\Rightarrow x=-4\)

b)

\((x-1)^2+3x(x-1)+7=(2x-1)^2+5(x-3)\)

\(\Leftrightarrow (x-1)^2+3x(x-1)+7=x^2+(x-1)^2+2x(x-1)+5(x-3)\)

\(\Leftrightarrow x(x-1)+7=x^2+5(x-3)\)

\(\Leftrightarrow 6x=22\Rightarrow x=\frac{11}{3}\)

c)

\(5(x^2-2x-1)+2(3x-2)=5(x+1)^2=5(x^2-2x+1)\)

\(\Leftrightarrow -5+2(3x-2)=5\)

\(\Leftrightarrow 3x-2=5\Rightarrow x=\frac{7}{3}\)

d)

\((x-1)(x^2+x+1)-2x=x(x-1)(x+1)=x(x^2-1)\)

\(\Leftrightarrow x^3-1-2x=x^3-x\Leftrightarrow -1-x=0\Rightarrow x=-1\)

3 tháng 10 2018

a,\((x+4)^2-(x+1)(x-1)=16\)

 \(\Rightarrow x^2+8x+16-x^2+1=16\)

\(\Rightarrow 8x=-1\Rightarrow x=-\dfrac{1}{8}\)

b,\((2x-1)^2-(x+3)^2-5(x+7)(x-7)=0\)

\(\Rightarrow 4x^2-4x+1-(x^2+6x+9)-5(x^2-49)=0\)

\(\Rightarrow 4x^2-4x+1-x^2-6x-9-5x^2-245=0\)

\(\Rightarrow -x^2-10x-244=0\)

\(\Rightarrow -(x^2-10x+25)-219=0\)

\(\Rightarrow -(x-5)^2-219=0\)

\(\Rightarrow (x-5)^2+219=0\)

Mà \((x-5)^2+219>0\) suy ra PT vô nghiệm

4 tháng 11 2018

\(\left(x+1\right)^2+\left(2x-1\right)\left(2x+1\right)-5x\left(x-1\right)=7\)

\(x^2+2x+1+4x^2-1-5x^2+5x=7\)

\(\left(x^2+4x^2-5x^2\right)+\left(1-1\right)+\left(2x+5x\right)=7\)

\(7x=7\)

\(x=1\)