Giải các phương trình
1,\(x\left(x-1\right)=2\left(x-1\right)\)
2, \(\left(x+2\right)\left(2x-3\right)=x^2-4\)
3, \(x^2+3x+2=0\)
4, \(5x^2+5x+3=0\)
5, \(x^3+x^2-12x=0\)
Bài 1: Giải phương trình
1) \(\sqrt{4x^2+12x+9}=2-x\left(vớix\le0\right)\)
2) \(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ( với \(x^2+5x+4>0\))
3) \(\sqrt{5x+1}=4\)
4) \(\sqrt{3-x}=7\)
Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\) ĐK:\(x\ge-1\)
\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)
\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)
Giải phương trình
\(x^4+x^3-12x^2=0\)
\(x^4-3x^3+3x^2-x=0\)
\(6x^4+5x^3-38x^2+5x+6=0\)
\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
x(x + 1). (x - 1 ). ( x + 2) = 24
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
giải phương trình
1)\(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
2)\(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
3)\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\)
4)\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}+\dfrac{3x-1}{x-4}-6\)
5)\(\dfrac{96}{x^2-16}=\dfrac{2x-1}{x+4}+\dfrac{3x-1}{x-4}-6\)
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
3)
\(\dfrac{x-3}{3}-\dfrac{x+2}{2}=\dfrac{x}{6}\\ \Leftrightarrow\dfrac{2\left(x-3\right)}{6}-\dfrac{3\left(x+2\right)}{6}=\dfrac{x}{6}\\ \Leftrightarrow2x-6-3x-6=x\\ \Leftrightarrow2x-3x-x=6+6\\ \Leftrightarrow-2x=12\\ \Leftrightarrow x=-6\)
Vậy PT có tập nghiệm S = { -6 }
Giair phương trình sau:
a,\(2x^3+5x^2-3x=0\) b,\(2x^3+6x^2=x^2+3x\)
c,\(x^2+\left(x+2\right)\left(11x-7\right)=4\) d,\(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
e, \(x^3+1=x\left(x+1\right)\) f,\(x^3+x^2+x+1=0\)
g,\(x^3-3x^2+3x-1=0\) h,\(x^3-7x+6=0\)
i,\(x^6-x^2=0\) j,\(x^3-12=13x\)
k,\(-x^5+4x^4=-12x^3\) l, \(x^3=4x\)
a) Ta có: \(2x^3+5x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)
\(\Leftrightarrow x\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
b) Ta có: \(2x^3+6x^2=x^2+3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{0;-3;\dfrac{1}{2}\right\}\)
c) Ta có: \(x^2+\left(x+2\right)\left(11x-7\right)=4\)
\(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x-18=0\)
\(\Leftrightarrow12x^2+24x-9x-18=0\)
\(\Leftrightarrow12x\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\12x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{3}{4}\right\}\)
Trong đó có nhiều phương trình kiến thức cơ bản mà nhỉ? Ít nâng cao, bạn lọc ra câu nào k làm đc thôi chứ!
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
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 \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Mn giúp mình vs
1, \(x^3-6x^2+10x-4=0\)
2, \(x^3+2x^2+2\sqrt{2}x+2\sqrt{2}=0\)
3, \(x^4+x^2-\sqrt{2}x+2=0
\)
4, \(x^4+5x^3-12x^2+5x+1=0\)
5, \(\left(x+5\right)\left(2x+12\right)\left(2x+20\right)\left(x+12\right)=3x^2\)
6, \(\left(x^2-5x+1\right)\left(x^2-4\right)=6\left(x-1\right)^2\)
7, \(x^4-9x^3+16x^2+18x+4=0\)
1. \(x^3-6x^2+10x-4=0\)
<=> \(\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
<=> \(\left(x-2\right)\left(x^2-4x+2\right)=0\)
<=> \(\orbr{\begin{cases}x=2\\x^2-4x+2=0\left(1\right)\end{cases}}\)
Giải pt (1): \(\Delta=\left(-4\right)^2-4.2=8>0\)
=> pt (1) có 2 nghiệm: \(x_1=\frac{4+\sqrt{8}}{2}=2+\sqrt{2}\)
\(x_2=\frac{4-\sqrt{8}}{2}=2-\sqrt{2}\)
1) Ta có: \(x^3-6x^2+10x-4=0\)
\(\Leftrightarrow\left(x^3-2x^2\right)-\left(4x^2-8x\right)+\left(2x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)-4x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-4x+2\right)\left(x-2\right)=0\)
+ \(x-2=0\)\(\Leftrightarrow\)\(x=2\)\(\left(TM\right)\)
+ \(x^2-4x+2=0\)\(\Leftrightarrow\)\(\left(x^2-4x+4\right)-2=0\)
\(\Leftrightarrow\)\(\left(x-2\right)^2=2\)
\(\Leftrightarrow\)\(x-2=\pm\sqrt{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2+\sqrt{2}\approx3,4142\left(TM\right)\\x=2-\sqrt{2}\approx0,5858\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,5858;2;3,4142\right\}\)
4) \(x^4+5x^3-12x^2+5x+1=0\)
<=> \(\left(x^4-x^3\right)+\left(6x^3-6x^2\right)-\left(6x^2-6x\right)-\left(x-1\right)=0\)
<=> \(\left(x^3+6x^2-6x-1\right)\left(x-1\right)=0\)
<=> \(\left[\left(x-1\right)\left(x^2+x+1\right)-6x\left(x-1\right)\right]\left(x-1\right)=0\)
<=> \(\left(x-1\right)^2\left(x^2-5x+1\right)=0\)
<=> \(\orbr{\begin{cases}x=1\\x^2-5x+1=0\left(1\right)\end{cases}}\)
Giải pt (1) ta có: \(\Delta=\left(-5\right)^2-4=21>0\)
=> pt có 2 nghiệm
\(x_1=\frac{5+\sqrt{21}}{2}\); \(x_2=\frac{5-\sqrt{21}}{2}\)
Bài Tập: Giải phương trình :
a) (x + 5)(2x - 3) = 0
b) \(\left(x^2-9\right)\left(4-x\right)=0\)
c) \(\left(2x+3\right)\left(4-5x\right)=0\)
d) \(2\left(x+3\right)\left(x-4\right)=0\)
e) \(\left(x^2-9\right)\left(4-x\right)=0\)
f) \(\left(2x+3\right)\left(x^2-16\right)=0\)
a: \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\5x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
=>x+3=0 hoặc x-4=0
=>x=-3 hoặc x=4
e: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=4\end{matrix}\right.\)
f: \(\Leftrightarrow\left(2x+3\right)\left(x-4\right)\left(x+4\right)=0\)
hay \(x\in\left\{-\dfrac{3}{2};4;-4\right\}\)
a, \(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{3}{2}\end{matrix}\right.\)
b, \(\Leftrightarrow\left[{}\begin{matrix}x^2-9=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm3\\x=4\end{matrix}\right.\)
c, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\4-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{4}{5}\end{matrix}\right.\)
d, \(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e, tương tự d
f, \(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\pm4\end{matrix}\right.\)