Tìm x
a) \(5x\left(3x-7\right)-15x\left(x-1\right)=3\)
b) \(\left(4x+2\right)\left(6x-3\right)-\left(8x+5\right)\left(3x-4\right)=2\)
c) \(7x^2-21x=0\)
d) \(9x^2-6x+1=0\)
e) \(16x^2-49=0\)
f) \(5x^3-20x=0\)
Tính:
\(a)\left(-2x^2\right)\cdot\left(3x-4x^3+7-x^2\right)\)
\(b)\left(x+3\right)\cdot\left(2x^2-3x-5\right)\)
\(c)\left(-6x^5+7x^4-6x^3\right):3x^3\)
\(d)\left(9x^2-4\right):\left(3x+2\right)\)
\(e)\left(2x^4-13x^3+15x^2+11x-3\right):\left(x^2-4x-3\right)\)
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
Tìm x biết
1) \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
2)\(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x+1\right)-33\)
3)\(6x\left(3x+5\right)-2x\left(9x-2\right)+\left(17-x\right)\left(x-1\right)+x\left(x-18\right)-17x^2=0\)
4)\(\left(x-1\right)\left(x+2\right)-\left(x-3\right)+5x-7=0\)
Giúp mình nha. Camon nhiều
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) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Giải các pt sau:
a) \(\sqrt{x+8}+\frac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)
b) \(x^4-2x^3+\sqrt{2x^3+x^2+2}-2=0\)
c) \(3x\sqrt[3]{x+7}\left(x+\sqrt[3]{x+7}\right)=7x^3+12x^2+5x-6\)
d) \(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
e) \(16x^2+19x+7+4\sqrt{-3x^2+5x+2}=\left(8x+2\right)\left(\sqrt{2-x}+2\sqrt{3x+1}\right)\)
f) \(\left(5x+8\right)\sqrt{2x-1}+7x\sqrt{x+3}=9x+8-\left(x+26\right)\sqrt{x-1}\)
g) \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
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!
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, \(6x\left(3-4x\right)+8x\left(3x-2\right)=16\)