tìm x : 2x(3x-5)=10x-6
Những câu hỏi liên quan
tìm x biết
a) 2/3 - 5/3x = 7/10x + 5/6
b) 2x-1/4=5/6-1/2x
Bài 2 : Tìm x , biết
a) ( 3x -1 ) (2x+7) - ( x +1) (6x-5 ) = 16
b) ( 10x +9 )x - ( 5x -1 ) (2x+3 )= 8
c) ( 3x - 5 ) ( 7- 5x ) + ( 5x +2 )( 3x-2 ) -2 = 0
d) x(x + 1) ( x+6 ) - x3 = 5x
24 tháng 10 2021 lúc 11:08
Tìm x biết
a). 3x-4/2x+5=3x+7/2x-20
b). 10x-5/7x+2=50x+10/35x-29
a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)
\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)
\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)
\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)
b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)
\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)
\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)
\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)
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rút gọn biểu thức sau:
B=\(\dfrac{x+5}{2x}-\dfrac{x-6}{5-x}-\dfrac{3x^2-2x-9}{2x^2-10x}\)(x≠0;x≠5)
\(=\dfrac{x+5}{2x}+\dfrac{x-6}{x-5}-\dfrac{3x^2-2x-9}{2x\left(x-5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x-5\right)+2x\left(x-6\right)-3x^2+2x+9}{2x\left(x-5\right)}\)
\(=\dfrac{x^2-25+2x^2-12x-3x^2+2x+9}{2x\left(x-5\right)}\)
\(=\dfrac{-10x-16}{2x\left(x-5\right)}=\dfrac{-5x-8}{x\left(x-5\right)}\)
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1. tìm x :
(x^2 - 1/2x) : 2x-(3x-1)^2 : (3x-1) = 0
2. tìm n thuộc N để phép chia là phép chia hết
a. (14x^5 - 7x^3 + 2x): 7x^n
b. (25x^7y^6 - 10x^5y^4 - 6x^3y^2): (-3x^ny^n)
23 tháng 7 2021 lúc 16:15
1) ( 3sqrt{50}-5sqrt{18}+3sqrt{8} ).sqrt{2}2. Tìm điều kiện:1)sqrt{-10x} 6) sqrt{dfrac{3x+21}{-5}}2) sqrt{6-3x} 7) sqrt{left(x^2+1right).2x}3) dfrac{-4}{sqrt{4x+6}} 8) sqrt{left(-x^2-2right)}.3x4) dfrac{5}{sqrt{2x}} 9) sqrt{dfrac{1}{left(x-1right)^2}}5) sqrt{dfrac{-1}{2x-6}}hộ mk với tí nx pk nộp r ;-; help
Đọc tiếp
1) ( \(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\) ).\(\sqrt{2}\)
2. Tìm điều kiện:
1)\(\sqrt{-10x}\) 6) \(\sqrt{\dfrac{3x+21}{-5}}\)
2) \(\sqrt{6-3x}\) 7) \(\sqrt{\left(x^2+1\right).2x}\)
3) \(\dfrac{-4}{\sqrt{4x+6}}\) 8) \(\sqrt{\left(-x^2-2\right)}.3x\)
4) \(\dfrac{5}{\sqrt{2x}}\) 9) \(\sqrt{\dfrac{1}{\left(x-1\right)^2}}\)
5) \(\sqrt{\dfrac{-1}{2x-6}}\)
hộ mk với tí nx pk nộp r ;-; help
Bài 1:
Ta có: \(\left(3\sqrt{50}-5\sqrt{18}+3\sqrt{8}\right)\cdot\sqrt{2}\)
\(=\left(15\sqrt{2}-15\sqrt{2}+6\sqrt{2}\right)\cdot\sqrt{2}\)
\(=6\sqrt{2}\cdot\sqrt{2}\)
=12
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Bài 2:
1) ĐKXĐ: \(x\le0\)
2) ĐKXĐ: \(x\le2\)
3) ĐKXĐ: \(x>\dfrac{-3}{2}\)
4) ĐKXĐ: x>0
5) ĐKXĐ: x<3
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Tìm x thuộc Z sao cho có giá trị nguyên
\(A=\frac{x^2-10x+36}{x-5}\)
\(B=\frac{2x^2+3x+3}{2x-1}\)
\(C=\frac{4x^2-6x-7}{x-3}\)
\(D=\frac{2x^2+x+6}{x+3}\)
\(E=\frac{3x^3+10x-5}{3x+1}\)
\(A=\frac{x^2-10x+36}{x-5}=\frac{x^2-10x+25+9}{x-5}\) \(=\frac{\left(x-5\right)^2+9}{x-5}=x-5+\frac{9}{x-5}\)
để \(A\in Z\)
<=> \(\frac{9}{x-5}\in Z\)mà \(x\in Z\)
=> \(x-5\inƯ\left(9\right)\)
=> \(x-5\in\left(1;-1;3;-3;9;-9\right)\)
=> \(x\in\left(6;4;8;2;14;-4\right)\)
học tốt
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1, tìm x biết:
a) 5x - 7( 2x - 5 ) < 2( x-1 )
b) 6 - 7( x - 4 ) lớn hơn hoặc bằng 3x + 2( 3- x )
c) 10x - 3( x - 5 )> 3x - 2( x - 4 )
thực hiện phép tính
\(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)
\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)
\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)
\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)
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c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)
\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)
\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)
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