√(2x - 2√(2x - 1)) - 2√(2x + 3 - 4√(2x - 1)) + 3√(2x + 8 - 6√(2x - 1)) = 4
Những câu hỏi liên quan
d) (3x – 5)(7 – 5x) – (5x + 2)(2 – 3x) = 4 g) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) =0 j) (2x – 1)(3x + 1) – (4 – 3x)(3 – 2x) = 3 k) (2x + 1)(x + 3) – (x – 5)(7 + 2x) = 8 m) 2(3x – 1)(2x + 5) – 6(2x – 1)(x + 2) = - 6
g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)
\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)
\(\Leftrightarrow14x=0\)
hay x=0
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Giải phương trình
\(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}}=4\)
Tìm x:
1) -3.(1-2x) - 4.(1+3x) = -5x + 5
2) 3.(2x - 5) - 6.(1 - 4x) = -3x + 7
3) (1 - 3x) - 2.(3x - 6) = -4x - 5
4) x.(4x - 3) - 2x.(2x - 1) = 5x - 7
5) 3x.(2x - 1) - 6x.(x + 2) = -3x + 4
6) (1 - 2x).3 - 4.(6x - 1) = 7x - 5
7) 6x - 3.(1 - 4x) - 5.(x + 1) = 2x + 7
8) 6.(1 - 3x) - 3.(2x + 5) = -10x + 7
9) 3x.(1 - 2x) + 6x^2 - 7x = 8.(1 - 2x) - 9
10) 2x.(1 + 3x) - 3x.(4 + 2x) = 3x - 4
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
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8 tháng 9 2021 lúc 14:22
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
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a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
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1) |2x - 1| 52) |2x - 1| |x + 5|3) |3x + 1| x - 2 4) |3 - 2x| x + 2 5) |2x - 1| 5 - x 6) |- 3x| x - 2 7) |2 - 3x| 2x + 1 8) |2x - 1| + |4x ^ 2 - 1| 0 9) (2x + 5)/(x + 3) + 1 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x) 10) (x - 1)/(x + 3) - x/(x - 3) (7x - 3)/(9 - x ^ 2) 11) 5 + 96/(x ^ 2 - 16) (2x - 1)/(x + 4) + (3x - 1)/(x - 4) 12) (2x)/(2x - 1) + x/(2x + 1) 1 + 4/((2x - 1)(2x + 1)) 13) (x + 2)/(x - 2) - 1/x 2/(x ^ 2 - 2x) 14) x/(2x - 6) + x/(2x + 2) (2x + 4)/(x ^ 2 - 2x - 3)
Đọc tiếp
1) |2x - 1| = 5
2) |2x - 1| = |x + 5|
3) |3x + 1| = x - 2
4) |3 - 2x| = x + 2
5) |2x - 1| = 5 - x
6) |- 3x| = x - 2
7) |2 - 3x| = 2x + 1
8) |2x - 1| + |4x ^ 2 - 1| = 0
9) (2x + 5)/(x + 3) + 1 = 4/(x ^ 2 + 2x - 3) - (3x - 1)/(1 - x)
10) (x - 1)/(x + 3) - x/(x - 3) = (7x - 3)/(9 - x ^ 2)
11) 5 + 96/(x ^ 2 - 16) = (2x - 1)/(x + 4) + (3x - 1)/(x - 4)
12) (2x)/(2x - 1) + x/(2x + 1) = 1 + 4/((2x - 1)(2x + 1))
13) (x + 2)/(x - 2) - 1/x = 2/(x ^ 2 - 2x)
14) x/(2x - 6) + x/(2x + 2) = (2x + 4)/(x ^ 2 - 2x - 3)
TÍnh nhanh: 1/2x(1+2) + 1/2x(1+2+3) + 1/2x(1+2+3+4) + ... + 1/2x(1+2+3+4+5+6+7+8+9)
TÍnh nhanh: 1/2x(1+2) + 1/2x(1+2+3) + 1/2x(1+2+3+4) + ... + 1/2x(1+2+3+4+5+6+7+8+9)
1) √(2x-1) <= 8-2x
2) √[(x+1)(4-x)] > x-2
3) √(x-2x^2+1) > 1-x
4) √(x+5) - √(x+4) > √(x+3)
5) √(5x-1) - √(x-1) > √(2x-4)
6) √(x+3) >= √(2x-8) + √(7-x)
7) √(x+2) - √(3-x) < √(5-2x)
8) √(x+1) > 3 - √(x+4)
9) √(5x-1) - √(4x-1)<= 3√x
10) { {√[2(x^2-16)]} / √(x-3) }+ √(x-3) > (7-x) / √(x-3)
Giúp mình 10 câu này với ạaa
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giải phương trình sau
1/ 2x( x+3) - 6 (x-3) =0
2/ 2x^2( 2x+3) +(2x+3) =0
3/ (x-2) (x+1) -(x-2) 4x =0
4/ 2x ( x-5) -3x +15=0
5/ 3x(x+4) -2x-8 =0
6/ x^2 (2x-6) + 2x -6 =0
1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
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4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
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