I 5-7x I =1/4
I 4x -11 I =1/2x-1
I x-5 I +I x-8 I=4-3x
Những câu hỏi liên quan
Bỏ dấu giá trị tuyệt đối
a) I - 2 - 1/2 x I với x < 2
b) I 4x - 1 I , 3x < 6
c) I - 5 - 3x I , 2x < 8
d ) I2 + 4/5 x I , 2x < 6
Giải phương trình:
1, 3x^2+6x-3sqrt{dfrac{x+7}{3}} (2 cách khác nhau )
2, left(sqrt{3x+1}-sqrt{x-2}right)left(sqrt{3x^2+7x+2}+4right)4x-2
3, sqrt{-3x-1}+sqrt{9x^2+9x+3}-9x^2-6x
4, sqrt{x^2+x-6}+3sqrt{x-1}sqrt{5x^2-1}
5, left(sqrt{x+4}+2right)left(x+2sqrt{x-5}+1right)6x
6, sqrt{5-x^4}-sqrt[3]{3x^2-2}1
7, 3x^2+11+sqrt{x-2}+sqrt{2x+3}14x
8, sqrt{x-sqrt{x-sqrt{x-sqrt{x-7}}}}7
9, sqrt{2x^2-1}+3xsqrt{x^2-1}3x^3+2x^2-9x-7 ( với x0 )
Đọc tiếp
Giải phương trình:
1, \(3x^2+6x-3=\sqrt{\dfrac{x+7}{3}}\) (2 cách khác nhau )
2, \(\left(\sqrt{3x+1}-\sqrt{x-2}\right)\left(\sqrt{3x^2+7x+2}+4\right)=4x-2\)
3, \(\sqrt{-3x-1}+\sqrt{9x^2+9x+3}=-9x^2-6x\)
4, \(\sqrt{x^2+x-6}+3\sqrt{x-1}=\sqrt{5x^2-1}\)
5, \(\left(\sqrt{x+4}+2\right)\left(x+2\sqrt{x-5}+1\right)=6x\)
6, \(\sqrt{5-x^4}-\sqrt[3]{3x^2-2}=1\)
7, \(3x^2+11+\sqrt{x-2}+\sqrt{2x+3}=14x\)
8, \(\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-7}}}}=7\)
9, \(\sqrt{2x^2-1}+3x\sqrt{x^2-1}=3x^3+2x^2-9x-7\) ( với \(x>0\) )
Bài 1: Giải phương trình sau:
a) left|frac{3x-6}{1-2x}right|x-2
b) left|-2x+8right|frac{x^2-6x+8}{x+3}
c) frac{left|xright|-6}{x^2-36}2
d) frac{left|x^2-4x+3right|}{5x^2-7x+2}x-3
e) frac{-2x^2+7x-4}{left|2x+1right|}4-x
f) frac{left|x^2+5x+4right|}{x^2+3x+2}x+4
Đọc tiếp
Bài 1: Giải phương trình sau:
a) \(\left|\frac{3x-6}{1-2x}\right|=x-2\)
b) \(\left|-2x+8\right|=\frac{x^2-6x+8}{x+3}\)
c) \(\frac{\left|x\right|-6}{x^2-36}=2\)
d) \(\frac{\left|x^2-4x+3\right|}{5x^2-7x+2}=x-3\)
e) \(\frac{-2x^2+7x-4}{\left|2x+1\right|}=4-x\)
f) \(\frac{\left|x^2+5x+4\right|}{x^2+3x+2}=x+4\)
a/ \(\left|\frac{3x-6}{1-2x}\right|=x-2\) \(\left(x\ne\frac{1}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{3x-6}{1-2x}=x-2\\\frac{3x-6}{1-2x}=2-x\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=\left(x-2\right)\left(1-2x\right)\\3x-6=\left(2-x\right)\left(1-2x\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}3x-6=x+4x-2-2x^2\\3x-6=-x-4x+2+2x^2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-2x^2+2x+4=0\\2x^2-8x+8=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
KL: .............
b/ Tương tự
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I 7+5x I = 1-4x
I 4x^2 - 2x I + 1 = 2x
I x^2 - 5x + 4 I = x+4
I 4 - 3x I = 3x -4
I 1+5x I = 1 + 5x
I x^2 - 3x + 1 I = 2x-3
I x-1 I = x^2 -x
|7 + 5x| = 1 - 4x
=> \(\orbr{\begin{cases}7+5x=1-4x\left(đk:x\le\frac{1}{4}\right)\\7+5x=4x-1\left(đk:x\ge\frac{1}{4}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}7-1=-4x-5x\\7+1=4x-5x\end{cases}}\)
=> \(\orbr{\begin{cases}6=-9x\\8=-x\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{2}{3}\left(tm\right)\\x=-8\left(ktm\right)\end{cases}}\)
|4x2 - 2x| + 1 = 2x
=> |4x2 - 2x| = 2x - 1
=> \(\orbr{\begin{cases}4x^2-2x=2x-1\left(đk:x\ge\frac{1}{2}\right)\\4x^2-2x=1-2x\left(đk:x\le\frac{1}{2}\right)\end{cases}}\)
=> \(\orbr{\begin{cases}4x^2-2x-2x+1=0\\4x^2-2x-1+2x=0\end{cases}}\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\4x^2-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\x^2=\frac{1}{4}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=\pm\frac{1}{2}\end{cases}}\)(tm)
Vậy ...
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Bài 1 : giải hệ phương trình
a.left{{}begin{matrix}x-3y112x+y1end{matrix}right.
b.left{{}begin{matrix}3x-2y75x-3y3end{matrix}right.
c.left{{}begin{matrix}x+2y12x+y2end{matrix}right.
Bài 2 : Giải phương trình
a.2x^2+3x-50 b.x^2-2x-70
c.x^2-4x-30 d.x^2-4x+20
Bài 3 : giải phương trình trùng phương
a.x^4-4x^2-50 b.2x^4-7x^2-40
c.x^4-7x^2-180 d.4x^4+x^2-...
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Bài 1 : giải hệ phương trình
a.\(\left\{{}\begin{matrix}x-3y=11\\2x+y=1\end{matrix}\right.\)
b.\(\left\{{}\begin{matrix}3x-2y=7\\5x-3y=3\end{matrix}\right.\)
c.\(\left\{{}\begin{matrix}x+2y=1\\2x+y=2\end{matrix}\right.\)
Bài 2 : Giải phương trình
a.\(2x^2+3x-5=0\) b.\(x^2-2x-7=0\)
c.\(x^2-4x-3=0\) d.\(x^2-4x+2=0\)
Bài 3 : giải phương trình trùng phương
a.\(x^4-4x^2-5=0\) b.\(2x^4-7x^2-4=0\)
c.\(x^4-7x^2-18=0\) d.\(4x^4+x^2-5=0\)
GIÚP MÌNH VỚI Ạ !!!!!!!!!
Giải phương trình:
1. (x - 4)2 - 25 0
2. (x - 3)2 - (x - 1)2 0
3. (x2 - 4)(2x +3) (x2 - 4)(x - 1)
4. (x2 - 1) - (x + 1)(2 - 3x) 0
5. x3 + x2 + x + 1 0
6. x3 + x2 - x - 1 0
7. 2x3 + 3x2 + 6x + 5 0
8. x4 - 4x3 - 19x2 + 106x - 120 0
9. (x2 - 3x + 2)(x2 + 15x + 56) + 8 0
Đọc tiếp
Giải phương trình:
1. (x - 4)2 - 25 = 0
2. (x - 3)2 - (x - 1)2 = 0
3. (x2 - 4)(2x +3) = (x2 - 4)(x - 1)
4. (x2 - 1) - (x + 1)(2 - 3x) = 0
5. x3 + x2 + x + 1 = 0
6. x3 + x2 - x - 1 = 0
7. 2x3 + 3x2 + 6x + 5 = 0
8. x4 - 4x3 - 19x2 + 106x - 120 = 0
9. (x2 - 3x + 2)(x2 + 15x + 56) + 8 = 0
1 ) \(\left(x-4\right)^2-25=0\)
\(\Leftrightarrow\left(x-4-5\right)\left(x-4+5\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-1\end{matrix}\right.\)
2 ) \(\left(x-3\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-1\right)\left(x-3-x+1\right)=0\)
\(\Leftrightarrow-2\left(2x-4\right)=0\)
\(\Leftrightarrow x=2.\)
3 ) \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x+3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=-4\end{matrix}\right.\)
4 ) \(\left(x^2-1\right)-\left(x+1\right)\left(2-3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1-2+3x\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{3}{4}\end{matrix}\right.\)
5 ) \(x^3+x^2+x+1=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-1.\end{matrix}\right.\)
6 ) \(x^3+x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
7 ) \(2x^3+3x^2+6x+5=0\)
\(\Leftrightarrow2x^3+2x^2+x^2+x+5x+5=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+x\left(x+1\right)+5\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x^2+x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=-1.\)
8 ) \(x^4-4x^3-19x^2+106x-120=0\)
\(\Leftrightarrow x^4-4x^3-19x^2+76x+30x-120=0\)
\(\Leftrightarrow x^3\left(x-4\right)-19x\left(x-4\right)+30\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-19x+30\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^3-8-19x+38\right)\left(x-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+4x+23\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
9 ) \(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x+7\right)\left(x+8\right)+8=0\)
\(\Leftrightarrow\left(x^2+7x-x-7\right)\left(x^2+8x-2x-16\right)+8=0\)
\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0\)
Đặt \(x^2+6x-7=t\)
\(\Leftrightarrow t\left(t-9\right)+8=0\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=8\\t=1\end{matrix}\right.\)
Khi t = 8 \(\Leftrightarrow x^2+6x-7=8\Leftrightarrow x^2+6x-15\Leftrightarrow\left[{}\begin{matrix}x=-3+2\sqrt{6}\\x=-3-2\sqrt{6}\end{matrix}\right.\)
Khi t = 1 \(\Leftrightarrow x^2+6x-7=1\Leftrightarrow x^2+6x-8=0\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{17}\\x=-3-\sqrt{17}\end{matrix}\right.\)
Vậy ........
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tim gia tri lon nhat cua bieu thuc :
a) C= 5+ 15/ 4 I 3x+7 I +3
b) D= 2 I 7x+5I +11/ I 7x+5I +4
tim GTNN cua bieu thuc :
a) A= I x+1I + 1,7
b) B= I x-2/3I +3/7
c) C= 5+ -8/ 4x I5x+7I 24
a)\([x.\dfrac{1}{2}]^{3}=\dfrac{1}{27}\)
b)\([x+\dfrac{1}{2} ]^{2}=\dfrac{4}{5} \)
c) I 3x-4/5 I = 11/5
d) I 2x - 2I = 0
\(a,\left(x.\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x.\dfrac{1}{2}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{3}:\dfrac{1}{2}=\dfrac{2}{3}\\ ---\\ b,\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{5}=\left(\dfrac{2}{\sqrt{5}}\right)^2=\left(-\dfrac{2}{\sqrt{5}}\right)^2 \\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\\x=-\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\end{matrix}\right.\\ Vậy:x=\pm\dfrac{2}{\sqrt{5}}-\dfrac{1}{2}\)
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\(c,\left|3x-\dfrac{4}{5}\right|=\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}=\dfrac{11}{5}\\3x-\dfrac{4}{5}=-\dfrac{11}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=\dfrac{11}{5}+\dfrac{4}{5}=3\\3x=-\dfrac{11}{5}+\dfrac{4}{5}=-\dfrac{7}{5}\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{3}=1\\x=-\dfrac{7}{5}:3=-\dfrac{7}{15}\end{matrix}\right.\\ ---\\ d,\left|2x-2\right|=0\\ \Leftrightarrow2x-2=0\\ \Leftrightarrow2x=2\\ \Leftrightarrow x=1\)
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a: (x*1/2)^3=1/27
=>x*1/2=1/3
=>x=1/3:1/2=2/3
b: \(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{2}{\sqrt{5}}\\x+\dfrac{1}{2}=-\dfrac{2}{\sqrt{5}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2\sqrt{5}}{5}-\dfrac{1}{2}=\dfrac{4\sqrt{5}-5}{10}\\x=\dfrac{-4\sqrt{5}-5}{10}\end{matrix}\right.\)
c: =>3x-4/5=11/5 hoặc 3x-4/5=-11/5
=>3x=3 hoặc 3x=-7/5
=>x=-7/15 hoặc x=1
d: =>2x-2=0
=>2x=2
=>x=1
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* Tìm x :
a, \(\left(3x-2\right)^2-\left(3x-5\right).\left(3x+2\right)=11\)
b, \(\left(4x-3\right)^2-\left(4x-5\right).\left(4x+5\right)=32\)
c, \(\left(5x-2\right)^2-\left(5x+3\right).\left(5x-5\right)=1\)
d, \(\left(x-4\right)^2-\left(x-7\right).\left(2x-3\right)=5-x^2\)
a) \(\left(3x-2\right)^2-\left(3x-5\right)\left(3x+2\right)=11\)
\(\Leftrightarrow\left(9x^2-12x+4\right)-\left(9x^2+6x-15x-10\right)=11\)
\(\Leftrightarrow9x^2-12x+4-9x^2-6x+15x+10=11\)
\(\Leftrightarrow-3x+3=0\)
\(\Leftrightarrow-3x=-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
b) \(\left(4x-3\right)^2-\left(4x-5\right)\left(4x+5\right)=32\)
\(\Leftrightarrow\left(16x^2-24x+9\right)-\left(16x^2-25\right)=32\)
\(\Leftrightarrow16x^2-24x+9-16x^2+25=32\)
\(\Leftrightarrow-24x+2=0\)
\(\Leftrightarrow-24x=-2\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy \(S=\left\{\dfrac{1}{12}\right\}\)
c) \(\left(5x-2\right)^2-\left(5x+3\right)\left(5x-5\right)=1\)
\(\Leftrightarrow\left(25x^2-20x+4\right)-\left(25x^2-25x+15x-15\right)=1\)
\(\Leftrightarrow25x^2-20x+4-25x^2+25x-15x+15=1\)
\(\Leftrightarrow-10x+18=0\)
\(\Leftrightarrow-10x=-18\)
\(\Leftrightarrow x=\dfrac{9}{5}\)
Vậy \(S=\left\{\dfrac{9}{5}\right\}\)
d) \(\left(x-4\right)^2-\left(x-7\right)\left(2x-3\right)=5-x^2\)
\(\Leftrightarrow\left(x^2-8x+16\right)-\left(2x^2-3x-14x+21\right)=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21=5-x^2\)
\(\Leftrightarrow x^2-8x+16-2x^2+3x+14x-21-5+x^2=0\)
\(\Leftrightarrow9x-10=0\)
\(\Leftrightarrow9x=10\)
\(\Leftrightarrow x=\dfrac{10}{9}\)
Vậy \(S=\left\{\dfrac{10}{9}\right\}\)
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