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Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
a) x + 15 = 36 - 2x
x + 15 = 36 - (x + x )
15 =36 - ( x + x) - x
15 = 36 - x - x - x
15 = 36 - 3x
3x = 36 - 15
3x = 21
x = 21 : 3
=> x = 7
b) (x - 7) - (2x +5) = -14
x - 7 -( 2x + 5) = -14
x - (2x + 5) = -14 + 7 = -7
x - 2x - 5 = -7
x - 2x = -7 + 5 = -2
x - x + x = 2
x = 2 (-x + x cũng bằng chính nó)
=> x = 2
c) (x - 12) - 15 = (-7 + 20) - (18+x)
(x - 12) - 15 = 13 - (18 + x)
(x - 12) - 15 = 13 - 18 - x
(x - 12) - 15 = -5 - x
15 = (x - 12 ) - (-5 - x)
15 = x - 12 + 5 + x
15 = x + (-12) + 5 + x
15 = 2x + [(-12) + 5]
15 = 2x + -7
2x = -7 + 15
2x = 8
x = 8 : 2
=> x = 4
..................
a) -5(x - 3) - 2(5 - 3x) = -(x - 1)
=> -5x + 15 - 10 + 6x = -x + 1
=> x + 5 = -x + 1
=> x + x = 1 - 5
=> 2x = -4
=> x = -4 : 2
=> x = -2
\(-5\left(x-3\right)-2\left(5-3x\right)=-\left(x-1\right)\)
\(\Leftrightarrow-5x+15-10+6x=-x+1\)
\(\Leftrightarrow x+5=1-x\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-4\div2\)
\(\Leftrightarrow x=-2\)
1:
a)10/20-15/20+16/20=-5/20+16/20
=11/20.
b)2/3+8/3:8/5=2/3+5/3
=7/3.
2:a)Ta có:
2x=1/4=3/4
2x=4/4=1
x=1:2
x=0,5
b)x:(2/12-1/12)=-3/8.
x:1/12=-3/8.
x=-3/8x1/12.
x=-1/32.
a) \(2\left(x-3\right)-3\left(x-1\right)=2x+6\)
\(\Leftrightarrow2x-6-3x+3=2x+6\)
\(\Leftrightarrow-3x=9\Leftrightarrow x=-3\)
Vậy x=-3
b) \(2\left(x+7\right)-4\left(x-2\right)=10\)
\(\Leftrightarrow2x+14-4x+8=10\)
\(\Leftrightarrow-2x=-12\Leftrightarrow x=-6\)
Vậy x=-6
c) \(10-\left(1-x\right)^2=6\Leftrightarrow\left(1-x\right)^2=4\)
\(\Leftrightarrow\left[\begin{matrix}1-x=-2\\1-x=2\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy x=3 hoặc x=-1
d)\(\left(x-3\right)^3-\left[\left(-2\right)^3+2^3\right]\left(1+3+5+7+...+2001\right)=-512\)
\(\Leftrightarrow\left(x-3\right)^3=-512\)
\(\Leftrightarrow x-3=-8\Leftrightarrow x=-5\)
Vậy x=-5
a) \(=12+43y-13-36x+105y-45+165x-285y+255\)
\(=-137y+209+129x\)
tương tự
bài 2
a) \(\left|16-3x\right|=-39+231\)
\(\left|16-3x\right|=192\)
đến đây xét 2 trường hợp
b) \(\left|6-2x\right|+5=3x-4\)
\(\left|6-2x\right|=3x-4-5\)
\(\left|6-2x\right|=3x-9\)
\(\Rightarrow\orbr{\begin{cases}6-2x=3x-9\\6-2x=9-3x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}-5x=-15\\x=3\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=3\end{cases}}\Rightarrow x=3\)
vậy...
mk làm mẫu mấy bài thôi, còn lại bạn suy nghĩ rồi làm
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
Nếu là tìm "cặp số nguyên" thì phải có x và y hoặc x với 1 chữ nào đấy. Bạn kiểm tra lại đề xem, chắc chỉ là "số nguyên x" thôi chứ?
\(x^2+3x+7⋮x+3\)
\(\Leftrightarrow xx+3x+7⋮x+3\)
\(\Leftrightarrow x\left(x+3\right)+7⋮x+3\)
Do \(x\left(x+3\right)⋮x+3\) nên \(7⋮x+3\)
\(\Leftrightarrow x+3\inƯ\left(7\right)=\left\{-1;1;-7;7\right\}\)
Ta có bảng sau:
\(x+3\) | \(-1\) | \(1\) | \(-7\) | \(7\) |
\(x\) | \(-4\) | \(-2\) | \(-10\) | \(4\) |
Vậy \(x\in\left\{-10;-4;-2;4\right\}\)
a: 18:(x-3)=6
=>x-3=18:6=3
=>x=3+3=6
b: \(3^{x-1}+4=85\)
=>\(3^{x-1}=81\)
=>\(3^{x-1}=3^4\)
=>x-1=4
=>x=5
c: 2x+3x=55
=>5x=55
=>x=55/5=11
d: \(\left(x-3\right)^3+11=53\)
=>\(\left(x-3\right)^3=53-11=42\)
=>\(x-3=\sqrt[3]{42}\)
=>\(x=\sqrt[3]{42}+3\)
a) 18 : (x - 3) = 6
x - 3 = 18 : 6
x - 3 = 3
x = 3 + 3
x = 6
b) 3ˣ⁻¹ + 4 = 85
3ˣ⁻¹ = 85 - 4
3ˣ⁻¹ = 81
3ˣ⁻¹ = 3⁴
x - 1 = 4
x = 4 + 1
x = 5
c) 2x + 3x = 55
5x = 55
x = 55 : 5
x = 11
d) Sửa đề:
(x - 3)³ - 11 = 53
(x - 3)³ = 53 + 11
(x - 3)³ = 64
(x - 3)³ = 4³
x - 3 = 4
x = 4 + 3
x = 7