Tìm x:
a) 27x3-27x2+9x-1=\(\dfrac{-1}{8}\)
b) x(4-x)+(2x-1)(x-4)=0
c) 3x(5x-2)-10x+4=0
Tìm x:
a) 5x(x-2)+(2-x)=0
b) x(2x-5)-10x+25=0
c) \(\dfrac{25}{16}\)-4x2+4x-1=0
d)x4+2x2-8=0
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a) \(5x\left(x-2\right)+\left(2-x\right)=0\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(x\left(2x-5\right)-10x+25=0\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(2x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{5}{2}\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(x-\dfrac{9}{8}\right)\left(x+\dfrac{1}{8}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{9}{8}=0\\x+\dfrac{1}{8}=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=-\dfrac{1}{8}\end{matrix}\right.\)
d) \(x^4+2x^2-8=0\)
\(\Rightarrow\left(x^4+2x^2+1\right)-9=0\)
\(\Rightarrow\left(x^2+1\right)^2-3^2=0\)
\(\Rightarrow\left(x^2+1-3\right)\left(x^2+1+3\right)=0\)
\(\Rightarrow\left(x^2-2\right)\left(x^2+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+4=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\end{matrix}\right.\) \(\Rightarrow x^2=2\) \(\Rightarrow x=\pm\sqrt{2}\)
Tìm x:
a)9x^2-30x+25=0
b)25x^2-5x+1/4=0
c)9x^2-25=0
d)(2x-1)^2-(3x+2)^2=0
a: \(9x^2-30x+25=0\)
\(\Leftrightarrow3x-5=0\)
hay \(x=\dfrac{5}{3}\)
c: \(9x^2-25=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)
b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)
c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)
\(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)
\(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
b) \(2x^3-5x^2+3x=0\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
a) \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(TH_1:3x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\)
\(TH_2:-2x-7=0\)
\(\Leftrightarrow x=-\dfrac{7}{2}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
b) \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(TH_1:x=0\)
\(TH_2:x-1=0\)
\(\Leftrightarrow x=1\)
\(TH_3:2x-3=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy pt có tập nghiệm \(S=\left\{0;1;\dfrac{3}{2}\right\}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(9x^2-16\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)-x\left(3x+4\right)=0\)
\(\Leftrightarrow\left(3x+4\right)\left(2x-4\right)=0\)
\(TH_1:3x+4=0\)
\(\Leftrightarrow x=-\dfrac{4}{3}\)
\(TH_2:2x-4=0\)
\(\Leftrightarrow x=2\)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{4}{3};2\right\}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Rightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x-9x=-6-16+12\)
\(\Leftrightarrow11x=-10\)
\(\Leftrightarrow x=-\dfrac{10}{11}\)
Vậy pt có nghiệm duy nhất \(x=-\dfrac{10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow3x+1=5x+8\)
\(\Leftrightarrow3x-5x=8-1\)
\(\Leftrightarrow-2x=7\)
\(\Leftrightarrow x=\dfrac{-7}{2}\)
Vậy \(X=\dfrac{-7}{2}\)
b) Ta có: \(2x^3-5x^2+3x=0\)
\(\Leftrightarrow x\left(2x^2-5x+3\right)=0\)
\(\Leftrightarrow x\left[\left(2x^2-2x\right)-\left(3x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x=1\) hoặc \(x=0\) hoặc \(x=\dfrac{3}{2}\)
c) \(9x^2-16-x\left(3x+4\right)=0\)
\(\Leftrightarrow9x^2-16-3x^2-4x=0\)
\(\Leftrightarrow6x^2-4x-16=0\)
\(\Leftrightarrow2\left(3x^2-2x-8\right)=0\)
\(\Leftrightarrow3x^2-6x+4x-8=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-4}{3}\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=\dfrac{-4}{3}\)
d) \(\dfrac{5x+4}{3}-1=\dfrac{3x-2}{4}\)
\(\Leftrightarrow\dfrac{20x+16}{12}-\dfrac{12}{12}=\dfrac{9x-6}{12}\)
\(\Leftrightarrow20x+16-12=9x-6\)
\(\Leftrightarrow20x+16-12-9x+6=0\)
\(\Leftrightarrow11x+10=0\)
\(\Leftrightarrow x=\dfrac{-10}{11}\)
Vậy \(x=\dfrac{-10}{11}\)
a) Ta có: \(9x^2-1=\left(3x-1\right)\left(5x+8\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)\left(5x+8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(3x+1-5x-8\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(-2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\-2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-2x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{1}{3};-\dfrac{7}{2}\right\}\)
1) (4-3x) (10x-5)=0
2) (7-2x) (4+8x) = 0
3) (9-7x) (11-3x) = 0
4) (7-14x) (x-2) = 0
5) (2x+1) (x-3) = 0
6) (8-3x) (-3x+5) = 0
7) (16-8x) (2-6x) = 0
8) (x+4) (6x-12) = 0
9) (11-33x) (x+11) = 0
10) (x-1/4) (x+5/6) = 0
11) (7/8-2x) (3x+1/3) = 0
12) 3x - 2x^2 = 0
13) 5x + 10x^2 = 0
14) 4x + 3x^2 = 0
15) -8x^2 + x =0
16) 10x^2 - 15x = 0
17) x^2 -4 =0
18) 9 - x^2 = 0
19) x^2 -1 = 0
20) (x-3) (2x-1) = (2x-1) ( 2x+3)
21) (5+4x) (-x+2) = (5+4x) (7+5x)
22) (4+x) (x-5) = (3x-8) (x-5) = 0
23) (3x-8) (7-21x) - (9+2x) (7-21x)
24) (10+ 7x) (x+1) = (9x-2)(x-1)
25) (9x-4) (x-1/2) - (x-1/2) (6+x) = 0
26) 9x^2 - 1 = (3x-1) (x+4)
27) (x+7) (3x+1) = 49-x^2
28) (2x+1)^2 = (x-1)^2
29)x^3- 5x^2+6x = 0
30) 3x^2 + 5x + 2 = 0
Giảii giúpp mìnhh đyy mọii ngườii .
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Tìm X:
a)(x-4)(x+4)=9
b)x2-4x+4-(5x-2)2=0
c)4x2+4+1-x2-10x-25=0
d)(x2+x+7)(x2+x-7)=(x2+x)2-7x
a)
⇔ \(x^2-16=9\)
⇔ \(x^2=25\)
⇔ \(x=\pm5\)
b)
⇔ \(x^2-4x+4-25x^2+20x-4=0\)
⇔ \(16x-24x^2=0\)
⇔ \(8x\left(2-3x\right)=0\)
⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)
c)
⇔ \(3x^2-10x-20=0\)
⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)
⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)
⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)
Vậy...
d)
⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)
⇔ 7x = 49
⇔ x=7
Vậy...
tìm x:
a)\(\dfrac{-3}{x+5}< 0\) b)\(\dfrac{2x+1}{7}< 0\) c)x\(^2\) - 5x + 4 >0 d)\(\dfrac{x+1}{x-1}< 1\)
a) 1\(\dfrac{2}{3}\). b)\(\dfrac{1}{7}\). c) 1 d )0
a: =>x+5>0
hay x>-5
b: =>2x+1<0
hay x<-1/2
c: =>(x-1)(x-4)>0
=>x>4 hoặc x<1
a) x>-5 ĐKXĐ x\(\ne\)-5
b)x<\(-\dfrac{1}{2}\)
c)x>4 hoặc x<1
d)ĐKXĐ x\(\ne\)1, ko tìm đc x
1).(4-3x)(10-5x)=0 2).(7-2x)(4+8x)=0 3).(9-7x)(11-3x)=0
4).(7-14x)(x-2)=0 5).(\(\dfrac{7}{8}\)-2x)(3x+\(\dfrac{1}{3}\))=0 6).3x-2x\(^2\)
7).5x+10x\(^2\)
1.
<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)
2.
<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3.
<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)
4.
<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)
5.
<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)
6,7. ko đủ điều kiện tìm
Tìm x:
a) x2+9x=0
b) (x+4)2-16=0
c) x3-16x=0
d) x2-10x+25=0
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
Tìm x:
a) x(2-x)+(x2+x)=7
b) (2x+1)2-x(4-5x)=17
c) (4-x)2-(2x+1)2=0
d) (2x3-8x2+10x) : (2x)=0
e) (4x4-16x-48) : (-2x)2=0
a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)
\(\Leftrightarrow x^2-4x+5=0\)
\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý)