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\(\frac{2}{7}\)x - \(\frac{1}{3}\)=\(\frac{3}{5}\)x-1
Tìm x biết:
5. ( x-1 ) - 7.( x-2 ) = 2x -39
Tìm x thuộc Z biết:
x - 3 - 14.( x-2 )= -3x -3
\(3x+7⋮x-2\)
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5 ( x - 1 ) - 7 ( x - 2 ) = 2x - 39
<=> 5x - 5 - 7x + 14 = 2x - 39
<=> 5x - 7x - 2x = -39 + 5 - 14
<=> -4x = -48
<=> x = 12
x - 3 - 14.( x-2 )= -3x -3\(\Rightarrow\chi-3-28-14\chi-28=-3\chi-3\)
\(\Rightarrow\chi-3-28+3=-3\chi-3\)
\(\Rightarrow\chi-28=11\chi\)
\(\Rightarrow\chi-11\chi=28\)
\(\Rightarrow10\chi=28\Rightarrow\chi=2,8\left(kot.m\chi\inℤ\right)\)
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\(3x^2-\left(x+2\right)\left(3x-1\right)=-7\)
\(\Rightarrow3x^2-\left(3x^2+6x-x-2\right)=-7\)
\(\Rightarrow3x^2-3x^2-5x+2=-7\)
\(\Rightarrow-5x+2=-7\)
\(\Rightarrow-5x=-9\)
\(\Rightarrow x=\frac{9}{5}\)
Vậy \(x=\frac{9}{5}\)
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1) x (x-2016) + 2015 (2016-x) = 0
x (x-2016) - 2015 (x- 2016) = 0
(x-2015)(x-2016) =0
\(\Rightarrow\orbr{\begin{cases}x-2015=0\\x-2016=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2015\\x=2016\end{cases}}}\)
Vậy x= 2015; 2016
2) -5x (x-15) + (15-x) = 0
-5x (x-15) - (x-15) =0
(-5x -1) (x-15) =0
\(\Rightarrow\orbr{\begin{cases}-5x-1=0\\x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}-5x=1\\x=15\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{5}\\x=15\end{cases}}}\)
Vậy x= -1/5; 15
3) 3x (3x-7) - (7-3x) =0
3x(3x-7) + (3x -7) =0
(3x+1) (3x-7) =0
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=-1\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{1}{3}\\x=\frac{7}{3}\end{cases}}}\)
Vậy x= -1/3 ; 7/3
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1,3x+2/5x+7 =3x-1/5x+1
<=> 1,3x+2/5x-3x+1/5x = 1-7
<=> (1,3+2/5-3+1/5)x = -6
<=> -11/10x=-6
<=> x= -6 : (-11/10)
<=> x= 60/11
2.x+1/2x+1 = 0,5x+2/x +3
<=> 2x+1/2x-0,5x-2/1x = 3-1
<=> x(2+1/2-0,5-2 ) =2
<=>0x =2
<=> x=0
Hinh nhu minh thay ban Kunzy Nguyen giai hoi sai
1,3x o dau ra ???????????????
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1) áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{3x+2}{5x+7}=\frac{3x-1}{5x+1}=\frac{\left(3x+2\right)-\left(3x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3x+2-3x+1}{5x+7-5x-1}=\frac{3}{6}=\frac{1}{2}\)
suy ra :
\(\frac{3x-1}{5x+1}=\frac{1}{2}\Rightarrow\left(5x+1\right).1=\left(3x-1\right).2\)
=> 5x+1=6x-2
5x-6x=-2-1
-x=-3
x=3
2)áp dụng tính chất của dãy tỉ số bằng nhau ta có;
\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}=\frac{\left(x+1\right)-2.\left(0,5x+2\right)}{\left(2x+1\right)-2.\left(x+3\right)}=\frac{x+1-x-4}{2x+1-2x-6}=\frac{-3}{-5}=\frac{3}{5}\)
suy ra:
\(\frac{x+1}{2x+1}=\frac{3}{5}\Rightarrow\left(2x+1\right).3=\left(x+1\right).5\)
=>6x+3=5x+5
6x-5x=5-3
x=2
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1)2.| x-1 |-3x=7
2|x-1|=7+3x
TH1:
2(x+1)=7+3x
2x+2=7+3x
2x-3x=7-2
-x=5
x=-5
TH2:
2[-(x-1)]=7+3x
2(-x+1)=7+3x
-2x+2=7+3x
-2x-3x=7-2
-5x=5
x=-1
vậy x={-1;-5}
2)Xác định hệ số sau để tổng các đơn thức sau bằng 5xy3, axy3,-3axy3,7xy3thiếu j nữa ko
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Ta có: \(2\left|x-1\right|-3x=7\)
\(\Leftrightarrow2\left|x-1\right|=7+3x\)
ĐKXĐ: \(7+3x\ge0\forall x\Leftrightarrow x\ge\frac{-7}{3}\)
Ta có: \(2\left|x-1\right|=7+3x\)
\(\Leftrightarrow\left|2x-2\right|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2=7+3x\\2x-2=-\left(7+3x\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2=7+3x\\2x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=7+2\\2x+3x=-7+2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=9\\5x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-9\left(l\right)\\x=-1\left(tm\right)\end{cases}}\)
Vậy: \(x=-1\)