Tìm x, biết :
3/1-2x = -5/3x-2
Những câu hỏi liên quan
Bài 2 Tìm x biết 1, (2x-2).(3x+1)-(3x-2).(2x-3)=5 2,(1-3x).(3x-5)-(2x-4)(2-3x)=x-6 3,(2x-1).(4x^2+2x+1)-(2x+1)(4x^2-2x+1)=5x+6 Giúp tớ với
Tìm x biết:
a) 3(2x-1)(3x-3)-(2x-1)(3x-3)=-3
b) (3x-1)(2x+7)-(x+1)(6x-5)=x+2-(x+5)
Tìm x biết
2(x - 3) + 5 = 3x - 1
2x(3x + 2) - 5 = 3( 2x^2 - 2x + 1)
(3x - 2)(2x - 3) + 5 = 5
giải chi tiết giùm mình
2(x - 3) + 5 = 3x - 1
2x-6+5=3x-1
2x-1=3x-1
2x-3x=-1+1
-x=0
x=0
2x(3x + 2) - 5 = 3( 2x^2 - 2x + 1)
6x2+4x-5=6x2-6x+3
6x2+4x-6x2+6x=3+5
10x=8
x=4/5
(3x - 2)(2x - 3) + 5 = 5
(3x-2)(2x-3)=0
=>3x-2=0 hoặc 2x-3=0
=>x=2/3 hoặc x=3/2
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2(x - 3) + 5 = 3x - 1
<=>2x-6+5=3x-1
<=>2x-3x=-1+6-5
<=>-x=0
<=>x=0
2x(3x + 2) - 5 = 3( 2x2 - 2x + 1)
<=>6x2+4x-5=6x2-6x+3
<=>4x+6x=3+5
<=>10x=8
<=>x=0,8
(3x - 2)(2x - 3) + 5 = 5
<=>(3x-2)(2x-3)=0
<=>3x-2=0 hoặc 2x-3=0
<=>x=2/3 hoặc x=3/2
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Tìm x biết:
a; 3x.(2x+3)-(2x+5x).(3x-2)=8
b;4.(x-1)-3.(x^2-5)_x^2=(x-3)-(x+4)
c; 2.(3x-1).(2x+5)-6.(2x-1).(x+2)=-6
d; 3.(2x-1).(3x-1)-(2x-3.(9x-1)-3=3
12 tháng 12 2021 lúc 14:43
Tìm x biết:
a) 2x(3x+1) – (2x+3)(3x-2) = 12
b) (x+2)2 – (x-3)(x+3) = 5
b: \(\Leftrightarrow4x+13=5\)
hay x=-2
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a) 2x(3x+1) – (2x+3)(3x-2) = 12
\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)
\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)
\(\Leftrightarrow-3x+6=12\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
vậy x = -2
b) (x+2)2 – (x-3)(x+3) = 5
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)
\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)
\(\Leftrightarrow4x+8=0\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy x = -2
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BT1: cho -3x(x+5)=-3x2-15x
(x+3)(x+2)=x2+5x+6
Tìm x biết:
--3x(x+5)+(x+3)(x+2)=7
BT2:Cho(2x+1)2=4x2+4x+1
(2x+1)(2x-1)=4x2-1
Tìm x biết:
(2x+1)2-(2x+1)(2x-1)=19
BT3: Tìm x biết:
a)x(x+1)-x(x+5)=9
b)4x2(x+5)-8x(x+7)=13
bài 1 :tìm x , biết :
(x-7)^ x+1(x-7)^x+11=0
bài 2 :tìm x , biết :
a,|2x-3| > 5 c,|3x-1| ≤ 7 d,|3x-5| + |2x+3| = 7
bài 3 :
a,tính tổng S = 1 + 5^2 + 5^4 + ....... + 5^200.
b,so sánh 2^30 + 3^30 + 4^30 và 3.24^10
Tìm x biết
1. 2(5x-8)-3(4x-5)=4(3x-4)+11
2. (2x+1)2-(4x-1).(x-3)-15=0
3. (3x-1).(2x-7)-(1-3x).(6x-5)=0
1) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)
\(\Rightarrow17x=17\Rightarrow x=1\)
3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)
\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
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2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)
\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)
\(\Leftrightarrow17x=17\)
hay x=1
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Tìm x biết:
1) 1/3x+2/5(x-1)=0
2) -2/3-1/3(2x-5)=3/2
3) (3x-1)[-1/2x+5]=0
Tìm x biết
A. 2.(3x-1)-5(x-3)-9(2x-4)=24
B. 2x2 +4( x2 -1)=2x (3x+ 1)
C. 2x (5-3x)+2x (3x-5)-3(x-7)=4
Đ. 5x(x+1)-4x(x+2=1-x
không ai trả lời
a,\(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)
\(< =>6x-2-5x+15-18x+36=24\)
\(< =>-29x+49=24< =>29x=25< =>x=\frac{25}{29}\)
b,\(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)
\(< =>2x^2+4x^2-4=6x^2+2x\)
\(< =>2x=-4< =>x=-\frac{4}{2}=-2\)
c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)
\(< =>10x-6x^2+6x^2-10x-3x+21=4\)
\(< =>-3x=4-21=-17< =>x=\frac{17}{3}\)
d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)
\(< =>5x^2+5x-4x^2-8x=1-x\)
\(< =>x^2-3x+x-1=0\)
\(< =>x^2-2x-1=0\)
\(< =>\left(x-1\right)^2=2\)
\(< =>\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{cases}}\)
Sai rồi bn:)
a, \(2\left(3x-1\right)-5\left(x-3\right)-9\left(2x-4\right)=24\)
\(\Leftrightarrow6x-2-5x+15-18x+36=24\)
\(\Leftrightarrow-17x+25=0\Leftrightarrow x=\frac{25}{17}\)
b, \(2x^2+4\left(x^2-1\right)=2x\left(3x+1\right)\)
\(\Leftrightarrow2x^2+4x^2-4=6x^2+2x\)
\(\Leftrightarrow-4-2x=0\Leftrightarrow x=-2\)
c, \(2x\left(5-3x\right)+2x\left(3x-5\right)-3\left(x-7\right)=4\)
\(\Leftrightarrow10x-6x^2+6x^2-10x-3x+21=4\)
\(\Leftrightarrow-3x+17=0\Leftrightarrow x=\frac{17}{3}\)
d, \(5x\left(x+1\right)-4x\left(x+2\right)=1-x\)
\(\Leftrightarrow5x^2+5x-4x^2-8x=1-x\)
\(\Leftrightarrow x^2-3x-1+x=0\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow x^2-2x+1-2=0\Leftrightarrow\left(x-1\right)^2=2\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}+1\\x=-\sqrt{2}+1\end{cases}}}\)
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