Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
e)
=> (x-2) . (x+7) = ( x-1 ) . ( x +4)
=> x2 +7x - 2x -14 = x2 - x + 4x - 4
x2 + 5x - 14 = x2 + 3x - 4
=> 5x - 14 = 3x - 4
=> 5x - 3x = 14-4
=> 2x = 10 => x = 10 : 2 => x = 5
c)
=>( x-1) . 7 = ( x + 5 ) . 6
=> 7x - 7 = 6x + 30
=> 7x - 6x= 30 + 7
=> x = 37
a,x=\(\frac{5}{2}\)
b,x=\(\frac{13}{176}\)
c,x=37
d, x=\(\frac{12}{5}\)
e, x=5
a: \(\Leftrightarrow x:\dfrac{3}{10}=\dfrac{33}{8}:\dfrac{1}{5}=\dfrac{165}{8}\)
\(\Leftrightarrow x=\dfrac{165}{8}\cdot\dfrac{3}{10}=\dfrac{99}{16}\)
b: \(\Leftrightarrow x\cdot\dfrac{1}{100}:4=\dfrac{11}{15}\)
\(\Leftrightarrow x\cdot\dfrac{1}{100}=\dfrac{44}{15}\)
hay x=880/3
chuyển vế bình hết lên ko thì xset 2 th mỗi th chắc dài lê thê nên ngại làm
GIẢI CHI TIẾT RA GIÚP MK VS NHÉ!!!!MK RẤT CẢM ƠN
a.
\(\left(x+\frac{1}{2}\right)\times\left(x-\frac{3}{4}\right)=0\)
TH1:
\(x+\frac{1}{2}=0\)
\(x=-\frac{1}{2}\)
TH2:
\(x-\frac{3}{4}=0\)
\(x=\frac{3}{4}\)
Vậy \(x=-\frac{1}{2}\) hoặc \(x=\frac{3}{4}\)
b.
\(\left(\frac{1}{2}x-3\right)\times\left(\frac{2}{3}x+\frac{1}{2}\right)=0\)
TH1:
\(\frac{1}{2}x-3=0\)
\(\frac{1}{2}x=3\)
\(x=3\div\frac{1}{2}\)
\(x=3\times2\)
\(x=6\)
TH2:
\(\frac{2}{3}x+\frac{1}{2}=0\)
\(\frac{2}{3}x=-\frac{1}{2}\)
\(x=-\frac{1}{2}\div\frac{2}{3}\)
\(x=-\frac{1}{2}\times\frac{3}{2}\)
\(x=-\frac{3}{4}\)
Vậy \(x=6\) hoặc \(x=-\frac{3}{4}\)
c.
\(\frac{2}{3}-\frac{1}{3}\times\left(x-\frac{3}{2}\right)-\frac{1}{2}\times\left(2x+1\right)=5\)
\(\frac{2}{3}-\frac{1}{3}x+\frac{1}{2}-x-\frac{1}{2}=5\)
\(\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}x+x\right)=5-\frac{2}{3}\)
\(-\frac{4}{3}x=\frac{13}{3}\)
\(x=\frac{13}{3}\div\left(-\frac{4}{3}\right)\)
\(x=\frac{13}{3}\times\left(-\frac{3}{4}\right)\)
\(x=-\frac{13}{4}\)
d.
\(4x-\left(x+\frac{1}{2}\right)=2x-\left(\frac{1}{2}-5\right)\)
\(4x-x-\frac{1}{2}=2x-\frac{1}{2}+5\)
\(4x-x-2x=\frac{1}{2}-\frac{1}{2}+5\)
\(x=5\)
a: \(\Leftrightarrow\left|x+2\right|=6x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{6}\\\left(6x+1-x-2\right)\left(6x+1+x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{6}\\\left(5x-1\right)\left(7x+3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{1}{5}\)
b: Trường hợp 1: x<2
Pt sẽ là 3-x+2-x=7
=>5-2x=7
=>2x=-2
hay x=-1(nhận)
Trường hợp 2: 2<=x<3
Pt sẽ là 3-x+x-2=7
=>1=7(vô lý)
Trường hợp 3: x>=3
Pt sẽ là x-3+x-2=7
=>2x-5=7
=>x=6(nhận)
d: \(\Leftrightarrow4^x\cdot\left(1+4^3\right)=4160\)
\(\Leftrightarrow4^x=64\)
hay x=3
\(\frac{x}{27}=-\frac{2}{3,6}\)
\(x=\frac{-2\times27}{3,6}\)
\(x=-15\)
***
\(\frac{-0,52}{x}=\frac{-9,36}{16,38}\)
\(x=\frac{-0,52\times16,38}{-9,36}\)
\(x=0,91\)
***
\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\frac{4,25}{2,875}=\frac{x}{1,61}\)
\(x=\frac{4,25\times1,61}{2,875}\)
\(x=2,38\)
\(a.\)
\(\frac{x}{27}=\frac{-2}{3,6}\)
\(\Rightarrow x=\frac{\left(-2\right).27}{3,6}\)
\(\Rightarrow x=\frac{-54}{3,6}\)
\(\Rightarrow x=-15\)
Vậy : \(x=-15\)
\(b.\)
\(-0,52:x=-9,36:16,38\)
\(\Rightarrow\frac{-0,52}{x}=-\frac{4}{7}\)
\(\Rightarrow x=\frac{-0,52.7}{-4}\)
\(\Rightarrow x=0,91\)
Vậy : \(x=0,91\)
\(c.\)
\(\frac{4\frac{1}{4}}{2\frac{7}{8}}=\frac{x}{1,61}\)
\(\Rightarrow\frac{\frac{17}{4}}{\frac{23}{8}}=\frac{x}{1,61}\)
\(\Rightarrow\frac{17}{4}.\frac{8}{23}=\frac{x}{1,61}\)
\(\Rightarrow\frac{34}{23}=\frac{x}{1,61}\)
\(\Rightarrow x=\frac{1,61.34}{23}\)
\(\Rightarrow x=\frac{119}{50}\)
Vậy : \(x=\frac{119}{50}\)
a) \(\frac{2}{\left(x+2\right).\left(x+4\right)}+\frac{4}{\left(x+4\right).\left(x+8\right)}+\frac{6}{\left(x+8\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14}{\left(x+2\right).\left(x+14\right)}-\frac{x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{x+14-x+2}{\left(x+2\right).\left(x+14\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow\frac{16}{\left(x+2\right).\left(x+4\right)}=\frac{x}{\left(x+2\right).\left(x+14\right)}\)
\(\Rightarrow x=16\)
Vậy x = 16
\(b,\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Leftrightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(vì\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)
\(\Leftrightarrow x=-1\)
\(\text{Vậy }x=-1\)
b. (x+1)(1/10+1/11+1/12-1/13-1/14)=0
x+1=0 (vì : 1/10+1/11+1/12-1/13-1/14>0)
x=-1
1)\(\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}=\frac{x+1}{5}+\frac{x+1}{6}\)
\(\Leftrightarrow\frac{x+1}{2}+\frac{x+1}{3}+\frac{x+1}{4}-\frac{x+1}{5}-\frac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0\).Do \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-\frac{1}{5}-\frac{1}{6}\ne0\)
\(\Leftrightarrow x=-1\)
anh ơi, câu 1 anh viết 0 . DO là ntn ạ ?