tìm x
a/ (x-5)^2-49=0
b/ (x+11)^2=121
c/ x.(x+7)-6x-42=0
d/ x^4-2x^3+10x^2-20x=0
bài 1 tìm x
a)6x^2-72x=0
b)-2x^4+16x=0
c)x(x-5)-(x-3)^2=0
d)(x-2)^3-(x-2)(x^2+2x+4)=0
a) \(6x^2-72x=0\)
\(6x\left(x-12\right)=0\)
\(6x=0\) hoặc \(x-72=0\)
*) \(6x=0\)
\(x=0\)
*) \(x-12=0\)
\(x=12\)
Vậy \(x=0;x=12\)
b) \(-2x^4+16x=0\)
\(-2x\left(x^3-8\right)=0\)
\(-2x=0\) hoặc \(x^3-8=0\)
*) \(-2x=0\)
\(x=0\)
*) \(x^3-8=0\)
\(x^3=8\)
\(x=2\)
Vậy \(x=0;x=2\)
c) \(x\left(x-5\right)-\left(x-3\right)^2=0\)
\(x^2-5x-x^2+6x-9=0\)
\(x-9=0\)
\(x=9\)
d) \(\left(x-2\right)^3-\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(x^3-6x^2+12x-8-x^3+8=0\)
\(-6x^2+12x=0\)
\(-6x\left(x-2\right)=0\)
\(-6x=0\) hoặc \(x-2=0\)
*) \(-6x=0\)
\(x=0\)
*) \(x-2=0\)
\(x=2\)
Vậy \(x=0;x=2\)
Tìm x , Biết
a) (x-4) x - (x-3)^2=0
b) 3x-6 = x^2-16
c) (2x-3)^2 - 49=0
d) 2x (x-5) - 7 (5-x)=0
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
tìm x
a 5x3-7x2-15x+21=0
b (x-3)2=4x2-20x+25
c x+x2-x3-x4=0
d 2x3+3x2+2x+3=0
b: 4x^2-20x+25=(x-3)^2
=>(2x-5)^2=(x-3)^2
=>(2x-5)^2-(x-3)^2=0
=>(2x-5-x+3)(2x-5+x-3)=0
=>(3x-8)(x-2)=0
=>x=8/3 hoặc x=2
c: x+x^2-x^3-x^4=0
=>x(x+1)-x^3(x+1)=0
=>(x+1)(x-x^3)=0
=>(x^3-x)(x+1)=0
=>x(x-1)(x+1)^2=0
=>\(x\in\left\{0;1;-1\right\}\)
d: 2x^3+3x^2+2x+3=0
=>x^2(2x+3)+(2x+3)=0
=>(2x+3)(x^2+1)=0
=>2x+3=0
=>x=-3/2
a: =>x^2(5x-7)-3(5x-7)=0
=>(5x-7)(x^2-3)=0
=>\(x\in\left\{\dfrac{7}{5};\sqrt{3};-\sqrt{3}\right\}\)
tim x
a) 4(2x+7)^2-9(x+3)^2=0
b) (5x^2-2x+10)^2=(3x^2+10x -8 )^2
c) (x-3)^2-4=0
d) x ^2-2x=24
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
B5:Giải pt:
a)2x\(^2\)-8=0
b)3x\(^3\)-5x=0
c)x\(^4\)+3x\(^2\)-4=0
d)3x\(^2\)+6x-9=0
e)\(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)
g)5x\(^4\)+6x\(^2\)-11=0
a. 2x\(^2\)-8=0
2x\(^2\)=8
x\(^2\)=4
x=2
b.3x\(^3\)-5x=0
x(3x\(^2\)-5)=0
\(\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}x=0\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=^+_-\sqrt{5}\end{matrix}\right.\)
c.x\(^4\)+3x\(^2\)-4=0\(^{\left(\cdot\right)}\)
đặt t=x\(^2\) (t>0)
ta có pt: t\(^2\)+3t-4=0 \(^{\left(1\right)}\)
thấy có a+b+c=1+3+(-4)=0 nên pt\(^{\left(1\right)}\) có 2 nghiệm
t\(_1\)=1; t\(_2\)=\(\dfrac{c}{a}\)=-4
khi t\(_1\)=1 thì x\(^2\)=1 ⇒x=\(^+_-\)1
khi t\(_2\)=-4 thì x\(^2\)=-4 ⇒ x=\(^+_-\)2
vậy pt đã cho có 4 nghiệm x=\(^+_-\)1; x=\(^+_-\)2
d)3x\(^2\)+6x-9=0
thấy có a+b+c= 3+6+(-9)=0 nên pt có 2 nghiệm
x\(_1\)=1; x\(_2\)=\(\dfrac{c}{a}=\dfrac{-9}{3}=-3\)
e. \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\) (ĐK: x#5; x#2 )
⇔\(\dfrac{\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}+\dfrac{3\left(x+2\right)\left(2-x\right)}{\left(x-5\right)\left(2-x\right)}\)=\(\dfrac{6\left(x-5\right)}{\left(x-5\right)\left(2-x\right)}\)
⇒2x - x\(^2\) + 4 - 2x + 6x - 6x\(^2\) + 12 - 6x - 6x +30 = 0
⇔-7x\(^2\) - 6x + 46=0
Δ'=b'\(^2\)-ac = (-3)\(^2\) - (-7)\(\times\)46= 9+53 = 62>0
\(\sqrt{\Delta'}=\sqrt{62}\)
vậy pt có 2 nghiệm phân biệt
x\(_1\)=\(\dfrac{-b'+\sqrt{\Delta'}}{a}=\dfrac{3+\sqrt{62}}{-7}\)
x\(_2\)=\(\dfrac{-b'-\sqrt{\Delta'}}{a}=\dfrac{3-\sqrt{62}}{-7}\)
vậy pt đã cho có 2 nghiệm x\(_1\)=.....;x\(_2\)=......
câu g làm tương tự câu c
Tìm x
a) 6x(5x + 3) + 3x(1 – 10x) = 7 b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
c) (x + 1)(x + 2)(x + 5) – x2(x + 8) = 27
d) 5x(12x + 7) – 3x(20x – 5) = - 100
e) 0,6x(x – 0,5) – 0,3x(2x + 1,3) = 0,138
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
d) 5x(12x + 7) – 3x(20x – 5) = - 100
⇒60x2+35x-60x2+15=-100
⇒35x+15=-100
⇒35x=-100-15
⇒35x=-115
⇒x=\(\dfrac{-115}{35}\)
⇒x=\(\dfrac{-23}{7}\)
1) x(x-5)-4x+20=0
2) 3(x+1)+x(x+1)
3) 2x^3+x=0
4) x^3-16x=0
5) x^2+6x=-9
6) x^4-2x^3+10x^2-20x=0
7) (2x-3)^2=(x+5)^2
1, x(x - 5) - 4x + 20 = 0
=> x(x - 5) - 4(x - 5) = 0
=> (x - 4)(x - 5) = 0
=> x - 4 = 0 hoặc x - 5 = 0
=> x = 4 hoặc x = 5
=> x thuộc {4; 5}
2, 3(x + 1) + x(x + 1)
= (3 + x)(x + 1)
3, 2x3 + x = 0
=> x(2x2 + 1) = 0
=> x = 0 hoặc 2x2 + 1 = 0
=> x = 0 hoặc 2x2 = -1
=> x = 0 hoặc x2 = -1/2 (vô lí vì x2 > hoặc = 0 với mọi x)
=> x = 0
4, x3 - 16x = 0
=> x(x2 - 16) = 0
=> x = 0 hoặc x2 - 16 = 0
=> x = 0 hoặc x2 = 16
=> x = 0 hoặc x = 4 hoặc x = -4
=> x thuộc {-4; 0; 4}
5, x2 + 6x = -9
=> x2 + 6x + 9 = 0
=> x2 + 2.3.x + 32 = 0
=> (x + 3)2 = 0
=> x + 3 = 0
=> x = -3
6, x4 - 2x3 + 10x2 - 20x = 0
=> x2(x2 + 10) - 2x(x2 + 10) = 0
=> (x2 + 2x)(x2 + 10) = 0
=> x(x +2)(x2 + 10) = 0
-TH1: x = 0
-TH2: x + 2 = 0 => x = -2
-TH3: x2 + 10 = 0 => x2 = -10 (vô lí vì x2 > hoặc = 0 với mọi x)
=> x thuộc {0; -2}
7, (2x - 3)2 = (x + 5)2
-TH1: 2x - 3 = x + 5
=> x = 8
- TH2: - 2x + 3 = x + 5
=> -3x = 2
=> x = \(\frac{-2}{3}\)
- TH3: 2x - 3 = - x - 5
=> 3x = -2
=> x = \(\frac{-2}{3}\)
- TH4: - 2x + 3 = - x - 5
=> -x = -8
=> x = 8`
=> x thuộc {\(\frac{-2}{3}\); 8}
Giải phương trình
a, (x^2-2)(x^2+x+1)=0
b, 16x^2 - 8x + 5=0
c, 2x^3 - x^2 - 8x + 4=0
d, 3x^3+6x^2 - 75x -150 = 0
e, 2x^5-3x^4+6x^3-8x^2+3=0
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
tim x
x^2-5x-4(x-5)=0
2x(x+6)=7x+42
x^3-5x^2+x-5=0
x^4-2x^3+10x^2-20x=0
(2x-3)-x^2+10x-25=0
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...