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TA
16 tháng 10 2019
1)3.x^2 - 75 = 0
3.x^2 - 3.25 = 0
3.(x^2-25)=0
x^2-5^2=0
(x-5)(x+5)=0
=> x-5=0 hoặc x+5=0
=> x=5 hoặc x=-5
KN
16 tháng 10 2019
1) \(3x^2-75=0\)
\(\Leftrightarrow3\left(x^2-25\right)=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)
2) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
3) \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1=1\)
\(\Leftrightarrow\left(x+1\right)^3=1^3\)
\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)
19 tháng 8 2017
\(x^2-36=0\Rightarrow x^2=36\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
\(3x^2-75=0\)
\(\Rightarrow3\left(x^2-25\right)=0\)
\(\Rightarrow x^2-25=0\Rightarrow x^2=25\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
\(4x^2-4x+1=0\)
\(\Rightarrow\left(2x-1\right)^2=0\)
\(\Rightarrow2x-1=0\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)
\(\left(x+3\right)^2-4=0\)
\(\Rightarrow\left(x+3\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\)
19 tháng 8 2017
a) \(x^2-36=0\Leftrightarrow x^2=36\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{36}\\x=-\sqrt{36}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
vậy \(x=6;x=-6\)
b) \(3x^2-75=0\Leftrightarrow3\left(x^2-25\right)=0\Leftrightarrow x^2-25=0\Leftrightarrow x^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{25}\\x=-\sqrt{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\) vậy \(x=5;x=-5\)
c) \(4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
d) \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x+3=\sqrt{4}\\x+3=-\sqrt{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=2\\x+3=-2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-5\end{matrix}\right.\) vậy \(x=-1;x=-5\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
7 tháng 11 2021
e) \(\left(9x^2-49\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\text{[}\left(3x\right)^2-7^2\text{]}+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x-7\right)\left(3x+7\right)+\left(3x+7\right)\left(7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\text{[}\left(3x-7\right)+\left(7x+3\right)\text{]}=0\)
\(\Rightarrow\left(3x+7\right)\left(3x-7+7x+3\right)=0\)
\(\Rightarrow\left(3x+7\right)\left(10x-4\right)=0\)
=> 2 TH
*3x+7=0 *10x-4=0
=>3x=-7 =>10x=4
=>x=-7/3 =>x=4/10=2/5
vậy x=-7/3 hoặc x=2/5
g) \(\left(x-4\right)^2=\left(2x-1\right)^2\)
\(\Rightarrow\left(x-4\right)^2-\left(2x-1\right)^2=0\)
\(\Rightarrow\left(x-4-2x+1\right)\left(x-4+2x-1\right)=0\)
\(\Rightarrow\left(-x-3\right)\left(3x-5\right)=0\)
\(\Rightarrow-\left(x+3\right)\left(3x-5\right)=0\)
=> 2 TH
*-(x+3)=0 *3x-5=0
=>-x=-3 =>3x=5
=x=3 =>x=5/3
h)\(x^2-x^2+x-1=0\)
\(\Rightarrow0+x-1=0\)
\(\Rightarrow x-1=0\)
=>x=0+1
=>x=1
vậy x=1
k, x(x+ 16) - 7x - 42 = 0
=>x^2+16x-7x-42=0
=>x^2+9x-42=0
vì x^2>0
do đó x^2+9x-42>0
nên o có gt nào của x t/m y/cầu đề bài
m)x^2+7x+12=0
=>x^2+3x++4x+12=0
=>x(x+3)+4(x+3)=0
=>(x+4).(x+3)=0
=>2 TH
=> *x+4=0
=>x=-4
vậy x=-4
*x+3=0
=>x=-3
vậy x=-3
n)x^2-7x+12=0
=>x^2-4x-3x+12=0
=>x(x-4)-3(x-4)=0
=>(x-3).(x-4)=0
=>2 TH
*x-3=0=>x=0+3=>x=3
*x-4=0=>x=0+4=>x=4
vậy x=3 hoặc x=4
KA
7 tháng 11 2021
a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1a)(3x−3)(5−21x)+(7x+4)(9x−5)=44⇔15x−63x2−15+63x+63x2−35x+36x−20=44⇔79x−35=44⇔79x=79⇒x=1
b)(x+1)(x+2)(x+5)−x2(x+8)=27⇔x2+2x+x+2(x+5)−x3−8x2=27⇔x2(x+5)+2x(x+5)+x(x+5)+2(x+5)−x3−8x2=27⇔x3+5x2+2x2+10x+x2+5x+2x+10−x3−8x2=27⇔17x+10=27⇔17x=17⇒x=1
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
31 tháng 7 2018
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
11 tháng 7 2018
\(x^3+x=0\)
\(\Rightarrow x.\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\end{cases}}\)
\(x^2-2x-3=0\)
\(\Rightarrow x.\left(x-2\right)=3\)
Vì \(x>x-2\)và \(x\inƯ\left(3\right)=\left\{3;-3\right\}\)
Các phần sau tương tự
KT
11 tháng 7 2018
\(x^3+x=0\)
\(\Leftrightarrow\)\(x\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
\(x^2-2x-3=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
\(x+5x^2=0\)
\(\Leftrightarrow\)\(x\left(5x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
Vậy...
1 tháng 9 2020
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
16 tháng 8 2017
ANH HAY CHỊ ƠI LÀM GIÚP EM BAI LỚP 7 ĐI O DUOI DAY A
16 tháng 8 2017
a) \(\left(x-3\right)^2-4=0\)
\(\Rightarrow\left(x-3\right)^2=4\)
\(\Rightarrow\left(x-3\right)^2=2^2=\left(-2\right)^2\)
\(\Rightarrow x-3=2\)hoặc \(\left(x-3\right)=-2\)
\(\Rightarrow\hept{\begin{cases}x-3=2\\x-3=-2\end{cases}\Rightarrow\hept{\begin{cases}x=5\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{5;-1\right\}\)
b) \(x^2-2x=24\)
\(\Rightarrow x.\left(x+2\right)=24\)
\(\Rightarrow x.\left(x+2\right)=4.6\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
3x2-75=0 3x2=0+75 3x2=75 x2=75:3 x2=25 x2=52 x=5 Vậy x=5
\(3x^2-75=0\)
\(3\left(x^2-25\right)=0\)
\(x^2-25^2=0\)
\(\left(x-5\right)\left(x+5\right)=0\)
\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)