- 2x*(x-7)+5x-35=0
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b,x2-2x-35=0
=>(x2-2x+1)-36=0
=>(x-1)2-62=0
=>(x-1-6)(x-1+6)=0
=>(x-7)(x+5)=0
=>x=7 hoặc x=-5
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
\(d,x\left(x-3\right)-7x+21=0\)
\(\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
\(a,2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
\(c,4x^2+12x+9=0\)
\(\Leftrightarrow4x^2+6x+6x+9=0\)
\(\Leftrightarrow2x\left(2x+3\right)+3\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)
\(b,5x+2\left(x-7\right)=35\\ \Leftrightarrow5x+2x-14-35=0\\ \Leftrightarrow7x-49=0\\ \Leftrightarrow7x=49\\ \Leftrightarrow x=7\\ d,đk:x\ne2;x\ne0\\ \dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x^2-2x}=0\\ \Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x\left(x+2\right)-\left(x-2\right)-2}{x\left(x-2\right)}=0\\ \Leftrightarrow x^2+2x-x+2-2=0\\ \Leftrightarrow x^2+x=0\\ \Leftrightarrow x\left(x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)
\(5x+2\left(x-7\right)=35\)
\(\Leftrightarrow5x+2x-14=35\)
\(\Leftrightarrow7x-14=35\)
\(\Leftrightarrow7x=49\)
\(\Leftrightarrow x=7\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{7\right\}\)
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x^2-2x}=0\) \(\text{ĐKXĐ:}x\ne0;x\ne2\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\)
\(\Rightarrow x^2+2x-x+2-2=0\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(\text{loại}\right)\\x=-1\left(\text{nhận}\right)\end{matrix}\right.\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{-1\right\}\)
a) \(21x+7=15x+35\)
\(\Leftrightarrow21x-15x=35-7\)
\(\Leftrightarrow6x=28\)
\(\Leftrightarrow x=\frac{28}{6}\)
b)\(|5x+3|-2x=x-17\)
\(\Leftrightarrow|5x+3|=3x-17\)
\(\Leftrightarrow\orbr{\begin{cases}5x+3=3x-17\\5x+3=17-3x\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-20\\8x=14\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x=-10\\x=\frac{4}{7}\end{cases}}\)
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-2\\3x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}}\)
a) \(21x+7=15x+35\)
\(\Leftrightarrow\)\(6x=28\)
\(\Leftrightarrow\)\(x=\frac{14}{3}\)
Vậy...
b) \(\left|5x+3\right|-2x=x-17\)
\(\Leftrightarrow\)\(\left|5x+3\right|=3x-17\)
Nếu \(x\ge-\frac{3}{5}\)thì pt trở thành:
\(5x+3=3x-17\)
\(\Leftrightarrow\)\(2x=-20\)
\(\Leftrightarrow\)\(x=-10\)(loại)
Nếu \(x< -\frac{3}{5}\)thì pt trở thành:
\(-5x-3=3x-17\)
\(\Leftrightarrow\)\(8x=14\)
\(\Leftrightarrow\) \(x=\frac{7}{4}\) (loại)
Vậy pt vô nghiệm
c) \(\left(5x+2\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x+2=0\\3x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{2}{5}\\x=\frac{4}{3}\end{cases}}\)
Vậy...
a) \(5x-65=5.3^2 \\ 5x-65=45\\5x=45+65\\5x=110\\x=22\)
b) \(200-(2x+6)=4^3\\2x+6=200-4^3\\2x+6=136\\2x=130\\x=65\)
c) \(2(x-51)=2.2^3+20\\2(x-51)=16+20\\2(x-51)=36\\x-51=18\\x=51+18=69\)
d) \(135-5(x+4)=35\\5(x+4)=135-45\\5(x-4)=90\\x-4=18\\x=18+4=22\)
e) \((2x-4)(15-3x)=0\\2(x-2).3(5-x)=0\\(x-2)(5-x)=0\\ \left[ \begin{array}{l}x-2=0\\5-x=0\end{array} \right. \\ \left[ \begin{array}{l}x=2\\x=5\end{array} \right.\)
f) \(2^{x+1} . 2^{2014}=2^{2016} \\ (2^{x+1} . 2^{2014}):2^{2014}=2^{2016} :2^{2014} \\ 2^{x=1}=2^{2016-2014} \\2^{x+1}=2^2\\x+1=2\\x=1\)
g) \(15+(x-1)^3=43\\(x-1)^3=15-42\\(x-1)^3=-27\\(x-1)^3=(-3)^3\\x-1=-3\\x=-2\)
h) \(15-x=17+(-9)\\15-x=17-9\\15-x=8\\x=15-8\\x=7\)
i) \(|x-5|=|-7|+|-4|\\|x-5|=7+4\\|x-5|=11\\ \left[ \begin{array}{l}x-5=11\\x-5=-11\end{array} \right. \\ \left[ \begin{array}{l}x=16\\x=-6\end{array} \right.\)
k) \(|x-3|-12=-9+|-7|\\|x-3|-12=-9+7\\|x-3|-12=-2\\|x-3|=10 \\ \left[ \begin{array}{l}x-3=10\\x-3=-10\end{array} \right. \\ \left[ \begin{array}{l}x=13\\x=-7\end{array} \right.\)
a) (x - 2)(x + 1) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
Vậy...
e) xy - 5x - 5y = 0
=> x(y - 5) - 5y = 0
=> x(y - 5) - 5(y - 5) - 25 = 0
=>(x - 5)(y - 5) = 25 = 1 . 25 = (-1) . (-25) = 5 . 5 = (-5). (-5) (và ngược lại)
Lập bảng :
x - 5 | 1 | 25 | -1 | -25 | 5 | -5 |
y - 5 | 25 | 1 | -25 | -1 | 5 | -5 |
x | 6 | 30 | 4 | -20 | 10 | 0 |
y | 30 | 6 | -20 | 4 | 10 | 0 |
Vậy ...
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
2x(x - 7) + 5x - 35 = 0
=> 2x(x - 7) + 5(x - 7) = 0
=> (x - 7)(2x + 5) = 0
=> x - 7 = 0 hoặc 2x + 5 = 0
=> x = 7 hoặc x = -5/2
Trả lời:
\(2x\left(x-7\right)+5x-35=0\)
\(\Leftrightarrow2x\left(x-7\right)+5\left(x-7\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\2x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-\frac{5}{2}\end{cases}}}\)
Vậy x = 7; x = - 5/2 là nghiệm của pt.