\(-\left(2x-1\right)\left(2x+1\right)\)
\(-\left(4x-5\right)\left(4x+5\right)\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
tìm x:
a) \(\left(2x+5\right)\left(4x-10\right)+4x.\left(3-2x\right)^2=0\)
b) \(\left(2x-3\right)^2-\left(3x+1\right)^2=0\)
a: =>8x^2-20x+20x-50+4x(4x^2-12x+9)=0
=>8x^2-50+8x^3-48x^2+36x=0
=>8x^3-40x^2+36x-50=0
=>\(x\simeq4,29\)
b: =>(2x-3-3x-1)(2x-3+3x+1)=0
=>(-x-4)(5x-2)=0
=>x=2/5 hoặc x=-4
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải các phương trình sau:
a)\(\left|4x+7\right|=2x+5\) b)\(\left|x^2-4x-5\right|=\left|4x-17\right|\)
c)\(\left|2x-5\right|+\left|2x^2-7x+5\right|\)=0 d)\(\left|x-1\right|+\left|2x+1\right|=\left|3x\right|\)
a/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+7=2x+5\\4x+7=-2x-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
b/ \(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=6\\x=\pm\sqrt{22}\end{matrix}\right.\)
c/ \(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\2x^2-7x+5=0\end{matrix}\right.\) \(\Rightarrow x=\frac{5}{2}\)
d/ \(\left|x-1\right|+\left|2x+1\right|\ge\left|x-1+2x+1\right|=\left|3x\right|\)
Dấu "=" xảy ra khi và chỉ khi: \(\left(x-1\right)\left(2x+1\right)\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)
Vậy nghiệm của pt là \(\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x-3\right)\left(2x+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
Sửa đề: \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2xy+5x-6y-15=2xy-2x+7y-7\\12xy-24x+3y-6=12xy+18x-2y-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x-6y-15=-2x+7y-7\\-24x+3y-6=18x-2y-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}7x-13y=8\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}42x-78y=48\\-42x+5y=3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}-73y=51\\7x-13y=8\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\7x=13y+8=-\dfrac{79}{73}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{51}{73}\\x=-\dfrac{79}{511}\end{matrix}\right.\)
Giải các PT:
a) \(\left(3x-2\right).\left(4x+5\right)=0\)
b) \(\left(2,3x-6,9\right).\left(0,1x+2\right)=0\)
c) \(\left(4x+2\right).\left(x^2+1\right)=0\)
d) \(\left(2x+7\right).\left(x-5\right).\left(5x+1\right)=0\)
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
Tìm số nguyên x
1/ \(17x+3\left(-16x-37\right)=2x+43-4x\)
2/ \(-2x-3\left(x-17\right)=34-2\left(-x+25\right)\)
3/ \(\left\{-3x+2\left[45-x-3\left(3x+7\right)-2x\right]+4x\right\}=55-103-57:\left[-2\left(2x-1\right)^2-\left(-9\right)^0\right]=-106\)
4/ \(-2x+3\left\{12-2\left[3x-\left(20+2x\right)-4x\right]+1\right\}=45\)
5/ \(3x-32>-5+1\)
6/ \(15+4x< 2x-145\)
7/ \(-3\left(2x+5\right)-16< -4\left(3-2x\right)\)
8/ \(-2x+15< 3x-7< 19-x\)
bài tập tết nâng cao phải ko
mk cũng có nhưng chưa làm dc
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, \(\left(3x+2\right).\left(6x-2\right)-\left(9x-2\right).\left(2x+1\right)=24\)
b, \(\left(4x+3\right)\left(3x-2\right)-\left(6x-1\right)\left(2x+3\right)=16\)
c, \(\left(5x-2\right)\left(4x+5\right)+\left(10x-7\right)\left(5-2x\right)=12\)
d, \(6x\left(3-4x\right)+8x\left(3x-2\right)=16\)