Tìm x biết :
a) \(3x^2+4x=0\)
b) \(25x^2-0,64=0\)
c) \(x^4-16x^2=0\)
d) \(x^2+x=6\)
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PQ
2
26 tháng 8 2021
a) 9-64x^2=0
=> 64x^2 = 8
=> \(x^2=\frac{8}{64}=\frac{1}{8}\)
=> \(x=\frac{1}{\sqrt{8}}\)
b ) 25x^2 - 3 = 0
=> 25x^2 = 3
=> \(x^2=\frac{3}{25}\)
=> \(x=\frac{\sqrt{3}}{5}\)
C) 7 - 16x^2 =0
=> 16x^2 = 7
=> \(x^2=\frac{7}{16}\)
=> \(x=\frac{\sqrt{7}}{4}\)
d) 4x^2 - (x-4)^2 = 0
=> 4x^2 - x^2 + 8x - 16 =0
=> 3x^2 + 8x -16 = 0
=> ( 3x^2 + 12x ) - ( 4x +16 ) = 0
=> 3x( x + 4 ) - 4( x + 4 ) = 0
=>( x + 4 )( 3x - 4 ) = 0
=> \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)
e) ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0
=> ( 3x + 4 + 2x - 5 )( 3x + 4 - 2x + 5 ) = 0
=> ( 5x -1 ) ( x + 9 ) = 0
=> \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)
QA
26 tháng 8 2021
Trả lời:
a, \(9-64x^2=0\)
\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)
Vậy x = 3/8; x = - 3/8 là nghiệm của pt.
b, \(25x^2-3=0\)
\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{3}}{5}\)
c, \(7-16x^2=0\)
\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{7}}{4}\)
d, \(4x^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(2x-x+4\right)\left(2x+x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)
Vậy x = - 4; x = 4/3 là nghiệm của pt.
e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+4-2x+5\right)\left(3x+4+2x-5\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = - 9; x = 1/5 là nghiệm của pt.
T
1
19 tháng 12 2021
a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
DV
1
1 tháng 11 2020
a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x^2\right)^2}=2\)
⇔ \(\left|x^2\right|=2\)
⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)
⇔ x2 - 2 = 0
⇔ ( x - √2 )( x + √2 ) = 0
⇔ x - √2 = 0 hoặc x + √2 = 0
⇔ x = ±√2
b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )
⇔ \(3\sqrt{x+1}=8\)
⇔ \(\sqrt{x+1}=\frac{8}{3}\)
⇔ \(x+1=\frac{64}{9}\)
⇔ \(x=\frac{55}{9}\)( tm )
c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )
⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)
⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)
⇔ \(7\sqrt{x-3}=14\)
⇔ \(\sqrt{x-3}=2\)
⇔ \(x-3=4\)
⇔ \(x=7\)( tm )
d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )
⇔ \(\left|3x-1\right|=5\)
⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x+2\right)^2}=6\)
⇔ \(\left|x+2\right|=6\)
⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)
1 tháng 11 2020
\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)
\(b)\)\(ĐK:x\ge0\)
\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)
Vậy \(x=\frac{64}{9}\)
\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)
- Nếu \(x\ge\frac{1}{3}\)thì \(\left(1\right)\Leftrightarrow3x-1=5\)\(\Leftrightarrow\)\(3x=6\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
- Nếu \(x< \frac{1}{3}\)thì \((1)\Leftrightarrow-\left(3x-1\right)=5\)\(\Leftrightarrow\)\(3x-1=-5\)\(\Leftrightarrow\)\(3x=-5+1\)\(\Leftrightarrow\)\(3x=-4\)\(\Leftrightarrow\)\(x=\frac{-4}{3}\left(TM\right)\)
Vậy \(x\in\hept{2;\frac{-4}{3}}\)
- \(e)\)\(\sqrt{x^2+4x+4}-6=0\)\(\Leftrightarrow\)\(\sqrt{(x+2)^2}=6\)\(\Leftrightarrow\)\(|x+2|=6\)\(\left(2\right)\)
-Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)
-Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)
Vậy \(x=4;x=-8\)
5 tháng 11 2017
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
30 tháng 9 2018
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
PT
1
15 tháng 2 2017
a) Gần giống cho nó giống luôn.
cần thêm (-x^3+2x^2-x) là giống
\(\left(x-1\right)^4+x^3-2x^2+x=\left(x-1\right)^4+x\left(x^2-2x+1\right)=\left(x-1\right)^4+x\left(x-1\right)^2\)
\(\left(x-1\right)^2\left[\left(x-1\right)^2+x\right]\)
\(\left[\begin{matrix}x-1=0\Rightarrow x=0\\\left(x-1\right)^2+x=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
Nghiệm duy nhất: x=1
Bài làm :
\(a\text{)}3x^2+4x=0\Leftrightarrow x\left(3x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{4}{3}\end{cases}}\)
\(b\text{)}25x^2-0,64=0\Leftrightarrow\left(5x-0,8\right)\left(5x+0,8\right)=0\Leftrightarrow\orbr{\begin{cases}5x-0,8=0\\5x+0,8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,16\\-0,16\end{cases}}\)
\(c\text{)}x^4-16x^2=0\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\Leftrightarrow\orbr{\begin{cases}x^2-4x=0\\x^2+4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x-4\right)=0\\x\left(x+4\right)=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
\(d\text{)}x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
Bài làm :
\(a)3x^2+4x=0\)
\(\Rightarrow x\left(3x+4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\3x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-4}{3}\end{cases}}\)
Vậy x = 0 hoặc \(x=\frac{-4}{3}\) .
\(b)25x^2-0,64=0\)
\(\Rightarrow\left(5x\right)^2=\frac{16}{25}\)
\(\Rightarrow\left(5x\right)^2=\left(\frac{4}{5}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{4}{5}\\5x=\frac{-4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{25}\\x=\frac{-4}{25}\end{cases}}\)
Vậy \(x=\frac{4}{25}\) hoặc \(x=\frac{-4}{25}\) .
\(c)x^4-16x^2=0\)
\(\Rightarrow x^2\left(x^2-16\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x^2-16=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=4^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy x = 0 hoặc \(x=\pm4\) .