tìm x biết :
a.\(5\left(x+3\right)-6x-2x^2=0\)
b.\(\left(x-2004\right)=8016x-4x^2\)
c.\(\left(x+1\right)^2=x+1\)
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`
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) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Tìm x, biết :
a/ \(\dfrac{1}{3}x\left(x^2-4\right)=0\)
b/ \(x\left(x+5\right)=x+5\)
c/ \(x^3-\dfrac{1}{9}x=0\)
3)\(^2-\left(x+5\right)^2=0\)
e/ \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
f/ \(x\left(2x-3\right)-6+4x=0\)
g/ \(2\left(3x-2\right)^2-9x^2+4=0\)
h/ \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
i/ \(4x^2+9x+5=0\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
A)\(^{ }\left(^{ }x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
B)\(^{ }\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
C)\(^{ }x^2-6x+11=0\)
D)(\(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
a) \(\left(x+1\right)\left(2x-1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+1=0\\2x-1=0\\-x+2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=\frac{1}{2}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{-1;\frac{1}{2};2\right\}\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}2x-1=0\\3x+2=0\\4x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\\x=\frac{5}{4}\\x=7\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{2}{3};\frac{5}{4};7\right\}\)
c) \(x^2-6x+11=0\)
\(\Leftrightarrow x^2-6x+9+2=0\)
\(\Leftrightarrow\left(x-3\right)^2+2=0\) (vô lí)
Vậy phương trình vô nghiệm
d) \(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left(x^2+2x+1+2\right)\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+5=0\\x-5=0\\x+19=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-5\\x=5\\x=-19\end{matrix}\right.\)
Vậy phương trình có tập nghiệm là \(S=\left\{\pm5;-19\right\}\)
a,b,d dễ mà bạn tự làm
c,x2-6x+11=0<=> x2-6x+9+2=0
<=>(x-3)2=-2(vô lý)
vậy pt vô nghiệm
Bài 1 :Rút gọn
\(\left(4x^2-3y\right)a2y-\left(3x^2-4y\right)3y\)
\(4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(2ax^2-a\left(1+2x^2\right)-\left\{a-x\left(x+a\right)\right\}\)
Bài 2:Tìm x
a)\(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+1=0\)
b)\(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
Bài 3:Rút gọn
\(x\left(1+x+x^2+...+x^9\right)-\left(1+x+x^2+...+x^9\right)\)
Bài 2 . Thực hiện phép tính
a)\(6x^3\)\(\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)\)\(-2x^5\)\(-x^3\)
b)\(\left(x-3\right)\left(x^2+3x-2\right)\)
c)\(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2
b: =x^3+3x^2-2x-3x^2-9x+6
=x^3-11x+6
c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)
\(=2x^2-3x-1+\dfrac{5}{2x+1}\)
a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)
\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)
\(=2x^5-16x^3-2x^5-x^3\)
\(=-17x^3\)
b) \(\left(x+3\right)\left(x^2+3x-2\right)\)
\(=x^3+3x^2-2x+3x^2+9x-6\)
\(=x^3+6x^2+7x-6\)
c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)
\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)
\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)
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,\(x^3-5x^2-14x=0\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
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