Tìm \(x\)
a, \(x^2-10x+25=0\)
b, \(x^2-8x+16=0\)
c, \(x^2-49=0\)
d, \(4x^2-25=0\)
Tìm x, biết:
a) 7x(x + 1) - 3(x + 1) =0
b) 3 ( x + 8) - x^2 - 8x = 0
c) x^2 - 10x = -25
d) x^2 - 10x = -25
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
Gi ải phương trình
a) \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\) b) \(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
c) \(\sqrt{x^2-10x+25}=2\) d) \(\sqrt{x^2-14x+49}-5=0\)
a: ĐKXĐ: x>=5
\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
b: ĐKXĐ: x>=1/2
\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)
=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)
=>\(5-\sqrt{2x-1}=0\)
=>\(\sqrt{2x-1}=5\)
=>2x-1=25
=>2x=26
=>x=13(nhận)
c: \(\sqrt{x^2-10x+25}=2\)
=>\(\sqrt{\left(x-5\right)^2}=2\)
=>\(\left|x-5\right|=2\)
=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
d: \(\sqrt{x^2-14x+49}-5=0\)
=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)
=>\(\sqrt{\left(x-7\right)^2}=5\)
=>|x-7|=5
=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)
\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)
\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)
Vậy \(S=\left\{9\right\}\)
\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)
Vậy \(S=\left\{13\right\}\)
\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
Vậy: \(S=\left\{7;3\right\}\)
\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{12;2\right\}.\)
Bài 1:Tìm x biết :
a) x3 - 0,25x = 0 c) x2 - 4x=0 d) 2x2 -4x=0
b) x2 - 10x = -25 d) x2 - 8x+16=0 e/ x2 -16 =0
a)\(x\left(x^2-0,25\right)=0\)
TH1:\(x=0\) TH2:\(x^2-0,25=0\)
\(x^2=0,25=>x=0,5\)
Vậy x E \(\hept{0,5;0}\)
Tìm x:
a) 18x^3-8x/25=0
b)x^3+2x^2-4x-8=0
c)x^2+10x+21=0
Tìm x thỏa mãn:
a, (x2 - 1)(x2 - 25) = 0
b, x2 - 8x + 16 = 0
c, x3 + 3x2 + 3x + 1 = 0
d, x3 + 10x2 + 25x = 0
a) \(\left(x^2-1\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\\x^2=25\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) \(x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
c) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Rightarrow x=-1\)
d) \(x^3+10x^2+25x=0\)
\(\Leftrightarrow x\left(x+5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
a) ( x2 - 1 )( x2 - 25 ) = 0
<=> \(\orbr{\begin{cases}x^2-1=0\\x^2-25=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm1\\x=\pm5\end{cases}}\)
b) x2 - 8x + 16 = 0
<=> ( x - 4 )2 = 0
<=> x - 4 = 0
<=> x = 4
c) x3 + 3x2 + 3x + 1 = 0
<=> ( x + 1 )3 = 0
<=> x + 1 = 0
<=> x = -1
d) x3 + 10x2 + 25x = 0
<=> x( x2 + 10x + 25 ) = 0
<=> x( x + 5 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Tìm x biết.
a) 4x^2 - 49 = 0 b) x^2 + 36 = 12x
c) 1/16x^2 - x + 4 = 0 d) x^3 -3√3x2 + 9x - 3√3 = 0
e) (x - 2)^2 - 16 = 0 f) x^2 - 5x - 14 = 0
g) 8x(x - 3) + x - 3 = 0
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
a,\(4x^2-49=0\)
\(\Leftrightarrow\left(2x\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)
b.\(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)
c.\(\frac{1}{16x^2}-x+4=0\)
\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)
\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)
........
Tim x,y biet:
1)x^2-2x+5+y^2-4y=0
2)4x^2+y^2-20x+26-2y=0
3)x^2+4y^2+13-6x-8y=0
4)4x^2+4x-6y+9x^2+2=0
5)x^2+y^2+6x-10y+34=0
6)25x^2-10x+9y^2-12y+5=0
7)x^2+9y^2-10x-12y+29=0
89x^2+12x+4y62+8y+8=0
9)4x^2+9y^2+20x-6y+26=0
10)3x^2+3y^2+6x-12y+15=0
11)x^2+4y^2+4x-4y+5=0
12)4x^2-12x+y^2-4y+13=0
13)x^2+y^2+2x-6y+10=0
14)4x^2+9y^2-4x+6y+2=0
15)y^2+2y+5-12x+9x^2=0
16)x^2+26+6y+9y^2-10x=0
17)10-6x+12y+9x^2+4y^2=0
18)16x^2+5+8x-4y+y^2=0
19)x^2+9y^2+4x+6y+5=0
20)5+9x^2+9y^2+6y-12x=0
21)x^2+20+9y62+8x-12y=0
22)x^2=4y+4y^2+26-10x=0
23)4y^2+34-10x+12y+x^2=0
24)-10x+y^2-8y+x^2+41=0
25)x^2+9y^2-12y+29-10x=0
26)9x^2+4y^2+4y+5-12x=0
27)4y^2-12x+12y+9x^2=13=0
28)4x^2+25-12x-8y+y^2=0
29)x62+17+4y^2+8x+4y=0
30)4y^2+12y+25+8x+x^2=0
31)x^2+20+9y^2+8x-12y=0
giup mk voi minh can gap ak, cam on cac ban
tìm x: a)x^4-2x^3+5x^2-10x=0
b)(3x+5)^2=(2x-2)^2
. c)x^3–2x^2+x=0
. d)x^2(x-1)-4x^2+8x-4=0
\(a,x^4-2x^3+5x^2-10x=0\\ \Leftrightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Leftrightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x^2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x\in\varnothing\left(x^2+5>0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
\(b,\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\\ \Leftrightarrow\left(3x+5+2x-2\right)\left(3x+5-2x+2\right)=0\\ \Leftrightarrow\left(5x+3\right)\left(x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-7\end{matrix}\right.\)
\(c,x^3-2x^2+x=0\\ \Leftrightarrow x\left(x-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(d,x^2\left(x-1\right)-4x^2+8x-4=0\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a) \(x^4-2x^3+5x^2-10x=0\\ \Rightarrow\left(x^4-2x^3\right)+\left(5x^2-10x\right)=0\\ \Rightarrow x^3\left(x-2\right)+5x\left(x-2\right)=0\\ \Rightarrow\left(x^3+5x\right)\left(x-2\right)=0\\ \Rightarrow x\left(x^2+5\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2+5=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\\x=2\end{matrix}\right.\)
Vậy \(x=\left\{-\sqrt{5};0;\sqrt{5};2\right\}\)
b) \(\left(3x+5\right)^2=\left(2x-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}3x+5=2x-2\\3x+5=-2x+2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
c) \(x^3-2x^2+x=0\\ \Rightarrow x\left(x^2-2x+1\right)=0\\ \Rightarrow x\left(x-1\right)^2=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
vậy ...
d) \(x^2\left(x-1\right)-4x^2+8x-4=0\\ x^2\left(x-1\right)-\left(4x^2-8x+4\right)=0\\ x^2\left(x-1\right)-\left(2x-2\right)^2=0\\ \Rightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\\ \Rightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Rightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-2\right)^2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a: Ta có: \(x^4-2x^3+5x^2-10x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+5x-10\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
b:Ta có: \(\left(3x+5\right)^2=\left(2x-2\right)^2\)
\(\Leftrightarrow\left(3x+5\right)^2-\left(2x-2\right)^2=0\)
\(\Leftrightarrow\left(3x+5-2x+2\right)\left(3x+5+2x-2\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Tìm x :
a) (x+2)^2-(x-2)(x+2)=0
b) (2x+3)^2-(x-1)^2=0
c) x^3-8=(x-2)^2
d) x^3+5x^2-4x-20=0
e) x^3-4x^2+4x=0
f) x^2-25+2(x+5)=0
g) 2(x^2+8x+16)-x^2+4=0
h) x^2(x-2)+7x=14
Mai nộp rồi giúp mình với các bạn