2x+x:6=12
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Mọi người check xem đúng không:x(3x+2)+(x+1)^2 - (2x - 5)(2x+5) -123x^2 + 2x + (x+1)^2 - 2x^2 - 5^2 -12 (3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x -12(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x -12 3x - 2x -12 ; 3x+2x -12 ; x+1+5 -12 ; x+1-5 -12 hoặc 2x -12 x -12 ; 5x -12 ; x+ 6 -12 ; x -4 -12 hoặc x -12: 2 x -12 ; x -12:5; x -12 :6; x -12 + 4 hoặc x -6 x -12 x -12/5; x -2 ; x -8 hoặc x -6
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Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
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Mọi người check xem đúng không:x(3x+2)+(x+1)^2 - (2x - 5)(2x+5) -123x^2 + 2x + (x+1)^2 - 2x^2 - 5^2 -12 (3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x -12(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x -12 3x - 2x -12 ; 3x+2x -12 ; x+1+5 -12 ; x+1-5 -12 hoặc 2x -12 x -12 ; 5x -12 ; x+ 6 -12 ; x -4 -12 hoặc x -12: 2 x -12 ; x -12:5; x -12 :6; x -12 + 4 hoặc x -6 x -12 x -12/5; x -2 ; x -8 hoặc x -6
Đọc tiếp
Mọi người check xem đúng không:
<=>x(3x+2)+(x+1)^2 - (2x - 5)(2x+5)= -12
<=>3x^2 + 2x + (x+1)^2 - 2x^2 - 5^2= -12
<=>(3x^2 - 2x^2) + [ (x +1)^2 -5^2] + 2x = -12
<=>(3x - 2x)(3x + 2x)+ (x+1-5) (x+1+5) + 2x = -12
=> 3x - 2x = -12 ; 3x+2x = -12 ; x+1+5 = -12 ; x+1-5 = -12 hoặc 2x = -12
=> x = -12 ; 5x =-12 ; x+ 6 = -12 ; x -4 = -12 hoặc x = -12: 2
=> x= -12 ; x = -12:5; x = -12 :6; x = -12 + 4 hoặc x= -6
=> x= -12 x = -12/5; x = -2 ; x = -8 hoặc x = -6
bạn làm sai rồi !
\(\Leftrightarrow x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=-12\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Leftrightarrow4x+26=-12\)
\(\Leftrightarrow4x=-38\)
\(\Leftrightarrow x=-\frac{19}{2}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{19}{2}\right\}\)
\(\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\\ =\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\\ =\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}=\dfrac{x^2-2x+6}{2x}\)
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\(\dfrac{x^2+2x}{2x+12}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+6\right)}+\dfrac{x-6}{x}+\dfrac{108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x\left(x^2+2x\right)+2\left(x+6\right)\left(x-6\right)+108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^3+2x^2+2\left(x^2-36\right)+108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-72+108-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^3+4x^2+36-6x}{2x\left(x+6\right)}\)
\(=\dfrac{x^3+6x^2-2x^2-12x+6x+36}{2x\left(x+6\right)}\)
\(=\dfrac{\left(x^3+6x^2\right)+\left(-2x^2-12x\right)+\left(6x+36\right)}{2x\left(x+6\right)}\)
\(=\dfrac{x^2\left(x+6\right)-2x\left(x+6\right)+6\left(x+6\right)}{2x\left(x+6\right)}\)
\(=\dfrac{\left(x+6\right)\left(x^2-2x+6\right)}{2x\left(x+6\right)}\)
\(=\dfrac{x^2-2x+6}{2x}\)
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1/ \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
2/ \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
3/ \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
4/ \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
5/ \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
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4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
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Tìm số nguyên x biết: a) x + 12 = 4; b) 19 - x = 0; c) 2x - 4 = -12;d) 2x- (-2) = 6
a)x=4-12=-8
b)x=19-0=19
c)x=-12+4:2=-4
d)x=6+(-2):2=2
1,Tìm x
5^x-5^x+2=650
2x-3/11+2x-4/12+2x-5/13=2x-6/14+2x-7/15+2x-8/6
5^x + 5^ ( x + 2 ) = 650
5x + 5x . 52 = 650
5x .( 1 + 25 ) = 650
5x . 26 = 650
5x = 650 : 26
5x = 25
5x = 52
=> x = 2
Vậy x = 2
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The Reflection Of Light~ Ủa đề là 5^x-5^x+2= 650 sao ông lại đổi thành 5^x-5^(x+2) zậy
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\(\frac{x-1}{12}+\frac{2x-12}{14}=\frac{3x-14}{25}+\frac{4x-29}{27}\)\(\frac{3}{x+1}+\frac{4}{x-2}=/frac{5}{x-3}+\frac{6}{x-4}\)\(\frac{2x-1}{2}+\frac{2x+1}{3}=/frac{2x+7}{6}+\frac{2x+9}{7}\)
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Cho 6 x - 12 y chia 9 = 8 z - 6 x 4 = 12 x - 8 chia 6 Tìm x y z biết 2x + z - y = 36
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\(A=\left(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}-\dfrac{2x^2}{3x^2-12}\right):\dfrac{6+13x}{24-12}\)
\(A=\left(\dfrac{2x+1}{2\left(x+2\right)}-\dfrac{x}{3\left(x-2\right)}-\dfrac{2x^2}{3\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{24-12x}{13x+6}\)
\(=\dfrac{3\left(2x+1\right)\left(x-2\right)-2x\left(x+2\right)-4x^2}{6\left(x-2\right)\left(x+2\right)}\cdot\dfrac{12\left(2-x\right)}{13x+6}\)
\(=\dfrac{3\left(2x^2-3x-2\right)-2x^2-4x-4x^2}{x+2}\cdot\dfrac{-2}{13x+6}\)
\(=\dfrac{6x^2-9x-6-6x^2-4x}{x+2}\cdot\dfrac{-2}{13x+6}=\dfrac{2}{x+2}\)
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GIẢI CÁC PT SAU:
\(\sqrt{5x+10}=8-x\)
\(\sqrt{4x^2+x-12}=3x-5\)
\(\sqrt{x^2-2x+6}=2x-3\)
\(\sqrt{3x^2-2x+6}+3-2x=0\)