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a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
\(\frac{2}{7}\)x - \(\frac{1}{3}\)=\(\frac{3}{5}\)x-1
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
\(5x-\frac{1}{3}=3x+\frac{2}{7}=8-\frac{5x}{2}\)
\(\Leftrightarrow5x-3x=\frac{2}{7}+\frac{1}{3}\)
\(\Leftrightarrow2x=\frac{13}{21}\)
\(\Leftrightarrow x=\frac{13}{42}\)
Thử lại:
\(5x-\frac{1}{3}=5\cdot\frac{13}{42}-\frac{1}{3}=\frac{17}{14}\)
\(3x+\frac{2}{7}=3\cdot\frac{13}{42}+\frac{2}{7}=\frac{17}{14}\)
\(8-\frac{5x}{2}=8-5\cdot\frac{13}{42}\div2=\frac{607}{84}\)( vô lý)
Vậy không có giá trị nào của x thoả mãn
2 ( 3x + 7 ) - 5 ( x - 4 ) = 0
=> 2 ( 3x + 7 ) = 5 ( x - 4 )
=> 6x + 14 = 5x - 20
=> x = -20 - 14
=> x = -34
Vậy x = -34
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 ???????????????
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
a: \(\Leftrightarrow12x^2-10x-12x^2-28x=7\)
=>-38x=7
hay x=-7/38
b: \(\Leftrightarrow-10x^2-5x+9x^2+6x+x^2-\dfrac{1}{2}x=0\)
=>1/2x=0
hay x=0
c: \(\Leftrightarrow18x^2-15x-18x^2-14x=15\)
=>-29x=15
hay x=-15/29
d: \(\Leftrightarrow x^2+2x-x-3=5\)
\(\Leftrightarrow x^2+x-8=0\)
\(\text{Δ}=1^2-4\cdot1\cdot\left(-8\right)=33>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-1-\sqrt{33}}{2}\\x_2=\dfrac{-1+\sqrt{33}}{2}\end{matrix}\right.\)
e: \(\Leftrightarrow-15x^2+10x-10x^2-5x-5x=4\)
\(\Leftrightarrow-25x^2=4\)
\(\Leftrightarrow x^2=-\dfrac{4}{25}\left(loại\right)\)
|5\(x\) - 4| = |\(x+2\)|
\(\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
vậy \(x\in\) { \(\dfrac{1}{3};\dfrac{3}{2}\)}
|2\(x\) - 3| - |3\(x\) + 2| = 0
|2\(x\) - 3| = | 3\(x\) + 2|
\(\left[{}\begin{matrix}2x-3=3x+2\\2x-3=-3x-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{5}\end{matrix}\right.\)
vậy \(x\in\){ -5; \(\dfrac{1}{5}\)}