Tìm x, biết :
a) \(\left|x-1\right|-2=x-3\)
b) \(\left|2x+3\right|+2x+5=2\)
c ) \(5x-\left|2x+5\right|=3x+5\)
d) \(3x+\left|5-2x\right|=10\)
Mỗi câu trên làm bằng 2 cách nhé !!!
giải phương trình
a.\(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
b.\(x\left(2x-9\right)=3x\left(x-5\right)\)
c.\(3x-15=2x\left(x-5\right)\)
d.\(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
e.\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
a) Ta có: \(\left(2x-3\right)^2=\left(2x-3\right)\left(x+1\right)\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x-3-x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=4\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};4\right\}\)
b) Ta có: \(x\left(2x-9\right)=3x\left(x-5\right)\)
\(\Leftrightarrow x\left(2x-9\right)-3x\left(x-5\right)=0\)
\(\Leftrightarrow x\left(2x-9\right)-x\left(3x-15\right)=0\)
\(\Leftrightarrow x\left(2x-9-3x+15\right)=0\)
\(\Leftrightarrow x\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
Vậy: S={0;6}
c) Ta có: \(3x-15=2x\left(x-5\right)\)
\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)
d) Ta có: \(\dfrac{5-x}{2}=\dfrac{3x-4}{6}\)
\(\Leftrightarrow6\left(5-x\right)=2\left(3x-4\right)\)
\(\Leftrightarrow30-6x=6x-8\)
\(\Leftrightarrow30-6x-6x+8=0\)
\(\Leftrightarrow-12x+38=0\)
\(\Leftrightarrow-12x=-38\)
\(\Leftrightarrow x=\dfrac{19}{6}\)
Vậy: \(S=\left\{\dfrac{19}{6}\right\}\)
e) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)
\(\Leftrightarrow6x+4-3x-1=12x+10\)
\(\Leftrightarrow3x+3-12x-10=0\)
\(\Leftrightarrow-9x-7=0\)
\(\Leftrightarrow-9x=7\)
\(\Leftrightarrow x=-\dfrac{7}{9}\)
Vậy: \(S=\left\{-\dfrac{7}{9}\right\}\)
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:
a. \(\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\)
b. \(-5\left(x+3\right)^2+\left(x-1\right)\left(x+1\right)+\left(2x-3\right)^2=\left(5x-2\right)^2-5x\left(5x+3\right)\)
\(a,\left(3x+4\right)\left(3x-4\right)-\left(2x+5\right)^2=\left(x-5\right)^2+\left(2x+1\right)^2-\left(x^2-2x\right)+\left(x-1\right)^2\\ \Leftrightarrow\left(9x^2-16\right)-\left(4x^2+20x+25\right)=x^2-10x+25+4x^2+4x+1-x^2+2x+x^2-2x+1\\ \Leftrightarrow9x^2-16-4x^2-20x-25=5x^2-6x+27\\ \Leftrightarrow5x^2-20x-41=5x^2-5x+27\\ \Leftrightarrow-15x=68\\ \Leftrightarrow x=-\dfrac{68}{15}\)Vậy..
Câu sau cũng tương tự nhé
giúp mk với tứ tư mk phải nộp rùi
bài 1:
a, \(2x\left(3x^2-5x+3\right)\)
b, \(-2x\left(x^2+5x-3\right)\)
c, \(\dfrac{-1}{2}x\left(2x^3-4x+3\right)\)
bài 2:
a,\(\left(2x-1\right).\left(x^2-5-4\right)\)
b,\(-\left(5x-4\right).\left(2x+3\right)\)
c,\(\left(2x-y\right).\left(4x^2-2xy+y^2\right)\)
d,\(\left(3x-4\right).\left(x+4\right).\left(5-x\right).\left(2x^2+3x-1\right)\)
e,\(7\left(x-4\right)-\left(7x+3\right).\left(2x^2-x+4\right)\)
bài 3:
c/m rằng gtri của biểu thức ko phụ thuộc vào gtri của biến
a,\(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
b,\(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
bài 4 :tìm x biết
a, \(3x+2\left(5-x\right)=0\)
b,\(x\left(2x-1\right).\left(x+5\right)-\left(2x^2+1\right).\left(x+4,5\right)=3,5\)
c,\(3x^2-3x\left(x-2\right)=36\)
d,\(\left(3x^2-x+1\right).\left(x-1\right)+x^2.\left(4-3x\right)=\dfrac{5}{2}\)
1,
a,\(2x\left(3x^2-5x+3\right)\)
\(=6x^3-10x^2+6x\)
b,\(-2x\left(x^2+5x-3\right)\)
\(=-2x^3-10x^2+6x\)
c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)
\(=-x^4+2x^2-\dfrac{3}{2}x\)
Bài 2:
a) \(\left(2x-1\right)\left(x^2-5-4\right)\)
\(=\left(2x-1\right)\left(x^2-9\right)\)
\(=2x^3-18x-x^2+9\)
b) \(-\left(5x-4\right)\left(2x+3\right)\)
\(=-\left(10x^2+15x-8x-12\right)\)
\(=-10x^2-7x+12\)
c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)
\(=8x^3-y^3\)
Bài 3: (chỉ cần kết quả ko chứa biến là ta có đpcm, nói chung bài này yêu cầu ta rút gọn)
a) \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)
\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)
\(=20\)
b) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)
\(=6x-3-5x+15+18x-24-19x\)
\(=-12\)
Tìm x, biết:
a) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
c) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)=4\)
\(\Leftrightarrow21x-15x^2-35+25x-4x+15x^2-4=4\)
\(\Leftrightarrow42x-39=4\)
\(\Leftrightarrow42x=4+39\)
\(\Leftrightarrow42x=43\)
\(\Rightarrow x=\dfrac{43}{42}\)
Tick cho mk nha
Giải các phương trình sau bằng cách đưa về phương trình tích :
a) \(\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
b) \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)
c) \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)
d) \(2x^3+5x^2-3x=0\)
a)(2x+1)(3x-2)=(5x-8)(2x+1)
⇔(2x+1)(3x-2)-(5x-8)(2x+1)=0
⇔(2x+1)(3x-2-5x+8)=0
⇔(2x+1)(-2x+6)=0
⇔2x+1=0 hoặc -2x+6=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-2x+6=0⇔-2x=-6⇔x=3
phương trình có 2 nghiệm x=-1/2 và x=3
b)4x2-1=(2x+1)(3x-5)
⇔(2x-1)(2x+1)-(2x+1)(3x-5)=0
⇔(2x+1)(2x-1-3x+5)=0
⇔(2x+1)(-x+4)=0
⇔2x+1=0 hoặc -x+4=0
1.2x+1=0⇔2x=-1⇔x=-1/2
2.-x+4=0⇔-x=-4⇔x=4
phương trình có 2 nghiệm x=-1/2 và x=4
Tìm x , biết :
a. \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
b. \(2x^3-50x=0\)
c.\(5x^2-4\left(x^2-2x+1\right)-5=0\)
d. \(x^3-x=0\)
e. \(27x^3-27x^2+9x-1=1\)
a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x+25=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
b) Ta có: \(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)
d) Ta có: \(x^3-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
e) Ta có: \(27x^3-27x^2+9x-1=1\)
\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)
\(\Leftrightarrow\left(3x-1\right)^3=1\)
\(\Leftrightarrow3x-1=1\)
\(\Leftrightarrow3x=2\)
hay \(x=\dfrac{2}{3}\)
Thực hiện phép tính:
a) \(2x.\left(2x^2+3x-1\right)\)
b) \(\left(x+5\right).\left(2x-3\right)\)
c) \(\left(x+1\right)^2-x\left(2+3x\right)\)
d) \(\left(2x^3+x^2-8x+3\right):\left(2x-3\right)\)
b: \(=2x^2-3x+10x-15=2x^2+7x-15\)
tìm x\(\in\) Q,biết:
a,\(\frac{x}{2}\left(\frac{3x}{5}\right)-\frac{13}{5}=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
b,\(\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
c,\(\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
d,\(\frac{13}{x-1}+\frac{5}{2x+2}-\frac{6}{3x-3}\)
e,\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-2\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
Ai bt câu nào làm câu đấy nhé ! Làm hết cũng đc nhé!