Giai BPT
a, ( 2 - x ) / 3x - 9 < hoặc = 0
b,(x + 2)^2 - (x - 3 ) ( x + 3 ) > 2x +5
Những câu hỏi liên quan
giải bpt
a. \(x^2-3x+2\le0\)
b.\(x^4-3x^2+2\le0\)
c.\(\dfrac{1}{x^2-x+1}\le\dfrac{1}{2x^2+x+2}\)
a: =>(x-1)(x-2)<=0
=>1<=x<=2
b: =>(x^2-1)(x^2-2)<=0
=>1<=x^2<=2
=>\(\left[{}\begin{matrix}1< =x< =\sqrt{2}\\-1>=x>=-\sqrt{2}\end{matrix}\right.\)
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Giai cac phuong trinh sau
1) (5x-4).(4x+6) 0
2) (4x-10).(24+5x) 0
3) (x-3).(2x+1) 0
4) (2x+1).(x2+2) 0
5) (x2+4).(7x-3) 0
6) (x-5).(3-2x).(3x+4) 0
7) (x-2).(3x+5) (2x-4).(x+1)
8) (2x+5).(x-4) (x-5).(4-x)
9) 9x2-1 (3x+1).(2x-3)
10) (2x-1)2 49
11) (5x-3)2 - (4x-7)2 0
12) (2x+7)2 9.(x+2)2
Giúp mình giải gấp .Cám ơn
Đọc tiếp
Giai cac phuong trinh sau
1) (5x-4).(4x+6) = 0
2) (4x-10).(24+5x) = 0
3) (x-3).(2x+1) = 0
4) (2x+1).(x2+2) = 0
5) (x2+4).(7x-3) = 0
6) (x-5).(3-2x).(3x+4) = 0
7) (x-2).(3x+5) = (2x-4).(x+1)
8) (2x+5).(x-4) = (x-5).(4-x)
9) 9x2-1 = (3x+1).(2x-3)
10) (2x-1)2 = 49
11) (5x-3)2 - (4x-7)2 = 0
12) (2x+7)2 = 9.(x+2)2
Giúp mình giải gấp .Cám ơn
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
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A giải các bất phương trình sau:
1, (x+4)/5 - x + 5 < (x+3)/3- (x-2)/2
2, (x+27)/5 - (3x-7)/4 >0
3, (7-8x)/(x^2+1) >0
4, (2x+1)/5 - (2x-2)/3 < 1
5, 1/(x+2) < 1/(x-2)
6, (x-2)/(x-5) - 3/(x-1) < 1
7, x + 6/x < 7
8, (3x-5)/x bé hơn hoặc bằng 2
9, (2x+1)/(x+1) bé hơn hoặc bằng 1
25 tháng 1 2021 lúc 21:36
Giai phường trình sau:a, 3x^2+2x-10 e, 4x^2-12x+50 i,2x^2+5x-30b,x^2-5x+60 f, 2x^2+5x+30 j,x^2+6x-160c,x^2-3x+20 g,x^2+x-20d,2x^2-6x+10 h, x^2-4x+30
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Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
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21 tháng 3 2023 lúc 20:09
Giai phương trình :
a)\(\dfrac{2x-1}{3}-x=\dfrac{x+3}{4}+2\)
b)\(x^2-4+\left(x-9\right)\left(x-2\right)=0\)
c)\(\dfrac{x-1}{x-3}-\dfrac{1}{x+3}=\dfrac{3x+3}{x^2-9}\)
a: =>4(2x-1)-12x=3(x+3)+24
=>8x-4-12x=3x+9+24
=>-4x-4=3x+33
=>-7x=37
=>x=-37/7
b: =>(x-2)(x+2+x-9)=0
=>(2x-7)(x-2)=0
=>x=2 hoặc x=7/2
c: =>(x-1)(x+3)-x+3=3x+3
=>x^2+2x-3-x+3=3x+3
=>x^2+x-3x-3=0
=>x^2-2x-3=0
=>(x-3)(x+1)=0
=>x=-1
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Giai phuong trinh:
c1. x^4+x^3-8x^2-9x-9=0
c2. x^4+2x^3-3x^2-8x-4=0
c3. x^4 +2x^3-3x^2-8x-4=0
Với dạng bài này ta chỉ việc chia hoocne là ra nhé!
\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)
\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
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Tìm x
a.(x+2).(x+3)-(x-2).(x+5) = 0
b. (2x+3).(x-4)+(x-5)(x+2) = (3x-5)(x-4)
c. (3x+2)(2x+9)-(x+2)(6x+1) = x+1-(x-6)
d. 3( 2x-1).(3x-1)-(2x-3).(9x-1)=0
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
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giai pt : a. x^4/2x^2+1 + 2x^2+1/x^4=2
b.(x/x-1)^2+(x/x+1)^2=10/9
c. x^3+3x^2-10x-24=0
giai bpt:
a) \(\frac{x-2}{4}+\frac{3x+4}{3}< 0\)
b) \(\frac{6x+9}{x-4}>0\)
c) \(\frac{2x-3}{2x+3}+\frac{2x+3}{2x-3}< 0\)
d) \(\frac{2x-3}{35}+\frac{x\left(x-2\right)}{7}\le\frac{x^2}{7}-\frac{2x-3}{5}\)