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\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
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a) 5(2x -1) - 4(8 - 3x) = 7
<=> 10x - 5 - 32 + 12x = 7
<=> 22x = 44
<=> x =2
Vậy x = 2 là nghiệm phương trình
b) 7(2x - 5) - 5(7x - 2) + 2(5x - 7) = (x - 2) - (x + 4)
<=> 14x - 35 - 35x + 10 + 10x - 14 = x - 2 - x - 4
<=> -11x - 39 = - 6
<=> -11x = 33
<=> x = -3
Vậy x = -3 là nghiệm phương trình
\(a,10x-5-32+12x=7\)
\(22x=44\)
\(x=2\)
\(b,14x-35-35x+10+10x-14=x-2-x-4\)
\(-11x-39=-6\)
\(-11x=-33\)
\(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)
<=> 10x - 35 + 16x - 10 = 5
<=> 10x + 16x = 5 + 35 + 10
<=> 26x = 50
<=> x = 50/26 = 25/13
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a) (5x-1)2-(5x-4)(5x+4)=7
\(25x^2-10x+1-\left(25x^2+20x-20x-16\right)=7\)
\(25x^2-10x+1-25x^2-20x+20x+16=7\)
\(-10x=7-1-16\)
\(-10x=-10\)
\(x=1\)
b) (x+5)2-x2=45
\(x^2+10x+25-x^2=45\)
\(10x=45-25\)
\(10x=20\)
\(x=2\)
Học tốt !!! có gì ko hiểu thì họi lại nhé em
a) (5x - 1)2 - (5x - 4)(5x + 4) = 7
25x2 - 10x + 1 - 25x2 + 16 = 7
-10x + 17 = 7
-10x = 7 - 17
-10x = -10
x = 1
b) (x + 5)2 - x2 = 45
x2 + 10x + 25 - x2 = 45
10x + 25 = 45
10x = 45 - 25
10x = 20
x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\Rightarrow5x^2-15x=5x^2-x-10x+2-5\)
\(\Rightarrow5x^2-15x-5x^2+x+10x=2-5\)
\(\Rightarrow-4x=-3\)
\(\Rightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
b) \(\Rightarrow x^2-4x-5x+20-x^2+2x-x+2=7\)
\(\Rightarrow x^2-4x-5x-x^2+2x-x=7-20-2\)
\(\Rightarrow-8x=-15\)
\(\Rightarrow x=\frac{15}{8}\)
Vậy \(x=\frac{15}{8}\)
c) \(\Rightarrow3x^2-6x-4x+8=3x^2-27x-3\)
\(\Rightarrow3x^2-6x-4x-3x^2+27x=-3-8\)
\(\Rightarrow17x=-11\)
\(\Rightarrow x=\frac{-11}{17}\)
Vậy \(x=\frac{-11}{17}\)
Chúc bạn học tốt.
\(a,\dfrac{4}{5}x-1=\dfrac{4}{7}\\ \dfrac{4}{5}x=\dfrac{4}{7}+1=\dfrac{11}{7}\\ x=\dfrac{11}{7}:\dfrac{4}{7}=\dfrac{11}{4}\)
\(b,\dfrac{2}{3}x+\dfrac{1}{2}x-1=-3\dfrac{1}{3}\\ \left(\dfrac{2}{3}+\dfrac{1}{2}\right)x-1=-\dfrac{10}{3}\\ \dfrac{7}{6}x=-\dfrac{10}{3}+1=-\dfrac{7}{3}\\ x=-\dfrac{7}{3}:\dfrac{7}{6}=-2\)
\(c,\left(4x^2-25\right)\left(2x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x^2-25=0\\2x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=\dfrac{25}{4}\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=\left(\pm\dfrac{5}{2}\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(d,\dfrac{2x+5}{-3}=\dfrac{-27}{2x+5}\\ \Leftrightarrow\left(2x+5\right)^2=\left(-27\right).\left(-3\right)=81=9^2=\left(-9\right)^2\\ \Leftrightarrow\left[{}\begin{matrix}2x+5=9\\2x+5=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\)
a: \(\dfrac{4}{5}x-1=\dfrac{4}{7}\)
=>\(\dfrac{4}{5}x=\dfrac{4}{7}+1=\dfrac{11}{7}\)
=>\(x=\dfrac{11}{7}:\dfrac{4}{5}=\dfrac{11}{7}\cdot\dfrac{5}{4}=\dfrac{55}{28}\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x-1=-3\dfrac{1}{3}\)
=>\(\dfrac{4}{6}x+\dfrac{3}{6}x=-\dfrac{10}{3}+1=-\dfrac{7}{3}\)
=>\(\dfrac{7}{6}x=-\dfrac{7}{3}\)
=>\(x=-\dfrac{7}{3}:\dfrac{7}{6}=-\dfrac{6}{3}=-2\)
c: \(\left(4x^2-25\right)\left(2x-3\right)=0\)
=>(2x-5)(2x+5)(2x-3)=0
=>\(\left[{}\begin{matrix}2x-5=0\\2x+5=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
d: \(\dfrac{2x+5}{-3}=\dfrac{-27}{2x+5}\)
=>\(\left(2x+5\right)^2=\left(-3\right)\cdot\left(-27\right)=81\)
=>\(\left[{}\begin{matrix}2x+5=9\\2x+5=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)