Giải phương trình sau:
a) x^3-x/49=0
b) x^2-7x+12=0
c)4x^2-3x-1=0
d) x^3-2x-4=0
giải các phương trình sau:
a.(x - 1)(x + 2)= 0
b.(x -2)(x -5)=0
c.(x +3)(x -5)=0
d.(x + 1/2)(4x + 4)=0
e.(x -4)(5x -10)=0
f.(2x -1)(3x +6)=0
g.(2,3x -6,9)(0,1x -2)=0
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
a.(x - 1)(x + 2)= 0
<=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
b.(x -2)(x -5)=0
<=> x-2=0 hoặc x-5=0
<=> x=2 hoặc x=5
c.(x +3)(x -5)=0
<=> x+3=0 hoặc x-5=0
<=> x=-3 hoặc x=5
d.(x + 1/2)(4x + 4)=0
<=> x+1/2=0 hoặc 4x+4=0
<=> x=-1/2 hoặc x=-1
e.(x -4)(5x -10)=0
<=> x-4=0 hoặc 5x-10=0
<=> x=4 hoặc x=2
f.(2x -1)(3x +6)=0
<=> 2x-1=0 hoặc 3x+6=0
<=> x=1/2 hoặc x=-2
g.(2,3x -6,9)(0,1x -2)=0
<=> 2,3x-6,9=0 hoặc 0,1x-2=0
<=> x=3 hoặc x=20
Bài 5: Giải các phương trình sau:
a. (3x - 1)2 - (x + 3)2 = 0
b. x3 = \(\dfrac{x}{49}\)
c. x2 - 7x + 12 = 0
d. 4x2 - 3x -1 = 0
e. x3 - 2x - 4 = 0
f. x3 + 8x2 + 17x +10 = 0
g. x3 + 3x2 + 6x + 4 = 0
h. x3 - 11x2 + 30x = 0
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
*Cách khác:
a) Ta có: \(\left(3x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)
Bài 2: giải phương trình sau
a) \(X^4\)-\(x^2\)-2=0
b) (x+1)\(^4\)-x\(^2\)+2)\(^2\)=0
c)3x\(^2\)-2x-8=0
Bài 3: giải phương trình sau
a) x\(^3\)-0,25=0
b) x\(^4\)+2x\(^3\)+x\(^2\)=0
c) x\(^3\)-1=0
d) 6x\(^2\)-7x+2=0
Mong có người giải giùm xin kẻm ơn :>
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
Giải các phương trình và hệ phương trình sau:
1. Phương trình bậc hai và hệ thức vi ét
a. -3² + 2x + 8=0
b. 5x² - 6x - 1=0
c. -3x² + 14x - 8=0
2. Nhẩm nghiệm của các phương trình bậc hai sau:
a) 5x² + 3x -2=0
b) -18x² + 7x +11=0
c) x² + 1001x + 1000 =0
d) -7x² - 8x + 15=0
e) 2x³ - 4x² - 6x =0
3. Tìm hai số biết tổng và tích của chúng:
a) u + v =14, uv=40
b) u + v = -7, uv=12
c) u + v = -5, uv = -24
3:
a: u+v=14 và uv=40
=>u,v là nghiệm của pt là x^2-14x+40=0
=>x=4 hoặc x=10
=>(u,v)=(4;10) hoặc (u,v)=(10;4)
b: u+v=-7 và uv=12
=>u,v là các nghiệm của pt:
x^2+7x+12=0
=>x=-3 hoặc x=-4
=>(u,v)=(-3;-4) hoặc (u,v)=(-4;-3)
c; u+v=-5 và uv=-24
=>u,v là các nghiệm của phương trình:
x^2+5x-24=0
=>x=-8 hoặc x=3
=>(u,v)=(-8;3) hoặc (u,v)=(3;-8)
Giải phương trình dạng tích:
a.16-25x^2=0
b.(x+1)^2-4=0
c.(3x+1)^2-4x^2=0
d.(4x+1)-(x-2)^2=0
e.(2x+1)^2-(x+3)^2=0
\(a,\Leftrightarrow\left(4-5x\right)\left(4+5x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1-2\right)\left(x+1+2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\\ c,\Leftrightarrow\left(3x+1-2x\right)\left(3x+1+2x\right)=0\\ \Leftrightarrow\left(x+1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{1}{5}\end{matrix}\right.\\ d,Sửa:\left(4x+1\right)^2-\left(x-2\right)^2=0\\ \Leftrightarrow\left(4x+1-x+2\right)\left(4x+1+x-2\right)=0\\ \Leftrightarrow\left(3x+3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{5}\end{matrix}\right.\\ e,\Leftrightarrow\left(2x+1-x-3\right)\left(2x+1+x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Giải các phương trình sau:
a/ (3x – 2)(4x + 5) = 0
b/ (2,3x – 6,9)(0,1x + 2) = 0
c/ (4x + 2)(x2 + 1) = 0
d/(2x + 7)(x – 5)(5x + 1) = 0
e/ (x – 1)(2x + 7)(x2 + 2) = 0
f/ (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
Bài 1: Giải các phương trình sau:
a) 3x ^ 2 - 5x + 2 = 0
d) - 4x ^ 2 + 25 = 0
b) 11x - 2x ^ 2 = 0
e) sqrt(x ^ 2 - x + 9) = 2x + 1
c) x ^ 2 + 5x + 7 = 0
f) 6x ^ 4 - 7x ^ 2 + 1 = 0
a: =>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
b: =>2x^2=11
=>x^2=11/2
=>\(x=\pm\dfrac{\sqrt{22}}{2}\)
c: Δ=5^2-4*1*7=25-28=-3<0
=>PTVN
f: =>6x^4-6x^2-x^2+1=0
=>(x^2-1)(6x^2-1)=0
=>x^2=1 hoặc x^2=1/6
=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)
d: =>(5-2x)(5+2x)=0
=>x=5/2 hoặc x=-5/2
e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2
=>3x^2+5x-8=0 và x>=-1/2
=>3x^2+8x-3x-8=0 và x>=-1/2
=>(3x+8)(x-1)=0 và x>=-1/2
=>x=1
Bài 2: Giải các phương trình sau:
a. (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b. x(x + 3)(x – 3) – 5(x + 2)(x2 – 2x + 4) = 0
c. x(x + 3)(x – 3) + 5(x – 3) = 0
d. (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1\right)=\left(3x-2\right)\left(3x+2\right)\left(x+1\right)\)
\(\Leftrightarrow x-1=3x-2\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c: =>x-3=0
hay x=3
d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
Bài 2: Giải các phương trình sau:
a. (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b. x(x + 3)(x – 3) – 5(x + 2)(x2 – 2x + 4) = 0
c. x(x + 3)(x – 3) + 5(x – 3) = 0
d. (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
\(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)
\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)
c: =>(x-3)(x2+3x+5)=0
=>x-3=0
hay x=3
d: =>(3x-1)(x2+2-7x+10)=0
=>(3x-1)(x-3)(x-4)=0
hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)
Phương trình nào tương đương với phương trình: x^2 -4 =0 ?
A. (2+x)(-x^2+2x+1)=0
B. x^2 -4x+4=0
C. (x-2)(x^2+3x+2)=0
D. căn x^2 -3 = 1