tìm x
\(4x^2+7x=0\)
\(48x-6x^4=0\)
\(\left(x-1\right)^3-x\left(x-2\right)^2=-1\)
Tìm x biết:
a) \(3x^2-4x=0\). b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\).
c) \(9x^2+6x+1=0\). d) \(x^2-4x=4\).
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Bài 3: Tìm x biết:
1, \(4x^2-36=0\)
2, \(\left(x-1\right)^2+x\left(4-x\right)=11\)
3, \(\left(x-5\right)^2-x.\left(x+2\right)=5\)
4, \(x\left(x+4\right)-x^2-6x=10\)
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
BÀI 6 tìm x
1,\(2x\left(x-5\right)-\left(3x+2x^2\right)=0\) 2,\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
3,\(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\) 4,\(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
5,\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\) 6,\(2x\left(1-x\right)+5=9-2x^2\)
1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)
=>-13x=0
=>x=0
2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
=>3x=13
=>x=13/3
3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
=>-2x^2=0
=>x=0
4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
=>-8x=6-14=-8
=>x=1
`1)2x(x-5)-(3x+2x^2)=0`
`<=>2x^2-10x-3x-2x^2=0`
`<=>-13x=0`
`<=>x=0`
___________________________________________________
`2)x(5-2x)+2x(x-1)=13`
`<=>5x-2x^2+2x^2-2x=13`
`<=>3x=13<=>x=13/3`
___________________________________________________
`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`
`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`
`<=>x=0`
___________________________________________________
`4)5x(x-1)-(x+2)(5x-7)=0`
`<=>5x^2-5x-5x^2+7x-10x+14=0`
`<=>-8x=-14`
`<=>x=7/4`
___________________________________________________
`5)6x^2-(2x-3)(3x+2)=1`
`<=>6x^2-6x^2-4x+9x+6=1`
`<=>5x=-5<=>x=-1`
___________________________________________________
`6)2x(1-x)+5=9-2x^2`
`<=>2x-2x^2+5=9-2x^2`
`<=>2x=4<=>x=2`
Giải phương trình:
*a) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
b) \(4x^4-5x^2+1=0\)
c) \(2x^4-7x^2+5=0\)
*d) \(x^4+7x^3-6x^2+7x+1=0\)
Viết mỗi tập hợp sau bằng cách liệt kê các phần tử:
a) A = { \(x\in R\) | \(\left(2x^2-5x+3\right)\left(x^2-4x+3\right)=0\) }
b) B = { \(x\in R\) | \(\left(x^2-10x+21\right)\left(x^3-x\right)=0\) }
c) C = { \(x\in R\) | \(\left(6x^2-7x+1\right)\left(x^2-5x+6\right)\) = 0 }
d) D = { \(x\in Z\) | \(2x^2-5x+3=0\) }
e) E = { \(x\in N\) | \(\left\{{}\begin{matrix}x+3< 4+2x\\5x-3< 4x-1\end{matrix}\right.\) }
f) F = { \(x\in Z\) | \(\left|x+2\right|\le1\) }
g) G = { \(x\in N\) | x < 5 }
h) H = { \(x\in R\) | \(x^2+x+3=0\) }
`a)(2x^2-5x+3)(x^2-4x+3)=0`
`<=>[(2x^2-5x+3=0),(x^2-4x+3=0):}<=>[(x=3/2),(x=1),(x=3):}`
`=>A={3/2;1;3}`
`b)(x^2-10x+21)(x^3-x)=0`
`<=>[(x^2-10x+21=0),(x^3-x=0):}<=>[(x=7),(x=3),(x=0),(x=+-1):}`
`=>B={0;+-1;3;7}`
`c)(6x^2-7x+1)(x^2-5x+6)=0`
`<=>[(6x^2-7x+1=0),(x^2-5x+6=0):}<=>[(x=1),(x=1/6),(x=2),(x=3):}`
`=>C={1;1/6;2;3}`
`d)2x^2-5x+3=0<=>[(x=1),(x=3/2):}` Mà `x in Z`
`=>D={1}`
`e){(x+3 < 4+2x),(5x-3 < 4x-1):}<=>{(x > -1),(x < 2):}<=>-1 < x < 2`
Mà `x in N`
`=>E={0;1}`
`f)|x+2| <= 1<=>-1 <= x+2 <= 1<=>-3 <= x <= -1`
Mà `x in Z`
`=>F={-3;-2;-1}`
`g)x < 5` Mà `x in N`
`=>G={0;1;2;3;4}`
`h)x^2+x+3=0` (Vô nghiệm)
`=>H=\emptyset`.
Giải các PT sau
a,\(\left(9^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
b,\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
c,\(\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
d,\(x^4+x^3+x+1=0\)
e,\(x^3-7x+6=0\)
f,\(x^4-4x^3+12x-9=0\)
g,\(x^5-5x^3+4x=0\)
h,\(x^4-4x^3+3x^2+4x-4=0\)
Tìm x, biết
a,\(\left(x^2+2x\right)^2-2x^2-4x=\)3
b,\(\left(x+\frac{1}{2}\right)^2-\left(x+\frac{1}{2}\right)\left(x+6\right)=8\)
c,\(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
d,\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-4\right)=1\)
e,\(3x^2+7x=10\)
g,\(\left(3x+5\right)\left(2x-1\right)-6x\left(x+2\right)=x\)
h,\(2\left(x+3\right)-x^2-3x=0\)
i,\(x^3-5x^2-14x=0\)
Tìm x
1) \(x^2-\left(x-1\right).\left(x+1\right)=0\)
2) \(5x^3+3x^2+3x+1=4x^3\)
3) \(x^3+x^2=0\)
4) \(2x^3-12x^2+18x=0\)
5)\(5x^2-4.\left(x^2-2x+1\right)+20=0\)
6)\(5x.\left(x-3\right)+7x-21=0\)
7)\(2x^3-50x=0\)
8) \(\left(4x-1\right)^2-9.\left(x+2\right)^2=0\)
9) \(3.\left(x-2\right)^2-x+2=0\)
10) \(9x^2+6x-8=0\)
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
giải phương trình, tiếp
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\)
\(\frac{1}{9}\left(x-3\right)^2-\frac{1}{25}\left(x+5\right)^2=0\)
\(2x^2-6x+1=0\)
\(3x^2+12x-66=0\)
\(9x^2-30x+225=0\)
\(3x^2-7x+1=0\)
\(3x^2-7x+8=0\)
\(x^2-4x+1=0\)
\(2x^2-6x+1=0\)
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
Bài 1. Giải các phương trình sau:
a.\(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
b. \(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
c. \(\left(x^2-1\right)\left(x+2\right)\left(x-3\right)=\left(x-1\right)\left(x^2-4\right)\left(x+5\right)\)
d. \(x^4+x^3+x+1=0\)
e. \(x^3-7x+6=0\)
h. \(x^4-4x^3+3x^2+4x-4=0\)
g. \(x^5-5x^3+4x=0\)
f. \(x^4-4x^3+12x-9=0\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
h.
\(x^4-4x^3+4x^2-x^2+4x-4=0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
g/
\(x\left(x^4-5x^2+4\right)=0\)
\(\Leftrightarrow x\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\\x=\pm2\end{matrix}\right.\)
f/
\(x^4-9-4x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)-4x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x^2-3\right)\left(x^2-4x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\pm\sqrt{3}\\x=1\\x=3\end{matrix}\right.\)