K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
G
4
21 tháng 11 2018
\(7^x+7^{x+2}=350\)
\(\Leftrightarrow7^x\left(1+7^2\right)=350\)
\(\Leftrightarrow7^x.50=350\)
\(\Leftrightarrow7^x=7\)
\(\Leftrightarrow7^x=7^1\)
\(\Leftrightarrow x=1\)
Vậy...
\(\left(2x-3\right)^2-1=35\)
\(\Leftrightarrow\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-3\right)^2=6^2\\\left(2x-3\right)^2=\left(-6\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
10 tháng 3 2017
\(1+\dfrac{1}{6}+\dfrac{120-x}{x}=\dfrac{120}{x}\)
\(1+\dfrac{1}{6}+\dfrac{126-\left(x+6\right)}{x+6}=\dfrac{120}{x}\)
\(1+\dfrac{1}{6}-1+\dfrac{126}{x+6}=\dfrac{120}{x}\)
\(\dfrac{1}{6}+\dfrac{126}{x+6}=\dfrac{120}{x}\)
\(\dfrac{126}{x+6}=\dfrac{120}{x}-\dfrac{1}{6}=\dfrac{120.6}{6x}-\dfrac{x}{6x}\)
\(\dfrac{126}{x+6}=\dfrac{126.6-x}{6x}\)
\(126.6.x=\left(126.6.-x\right)\left(x+6\right)\)ok
10 tháng 3 2017
đk: x khác -6 ,làm toán là khôn khéo, bn tim msc vế trái =6(x+6)
có: (6(x+6) + (x+6) + 6(120-x)) /6(x+6) = 120/x
bây gio bn rut gon r cho tich trung tỷ = ngoai ty la tim dc x
TT
2
KT
27 tháng 3 2018
\(\frac{x-144}{10}+\frac{x-130}{12}+\frac{x-112}{14}+\frac{x-106}{16}+\frac{x-96}{17}=15\)
\(\Leftrightarrow\)\(\frac{x-144}{10}-1+\frac{x-130}{12}-2+\frac{x-112}{14}-3+\frac{x-106}{16}-4+\frac{x-96}{17}-5=0\)
\(\Leftrightarrow\)\(\frac{x-154}{10}+\frac{x-154}{12}+\frac{x-154}{14}+\frac{x-154}{16}+\frac{x-154}{17}=0\)
\(\Leftrightarrow\)\(\left(x-154\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
\(\Leftrightarrow\)\(x-154=0\) (do 1/10 + 1/12 + 1/14 + 1/16 + 1/17 khác 0)
\(\Leftrightarrow\)\(x=154\)
Vậy...
NT
1
D
6 tháng 8 2019
\(ĐK:x\ne0;x\ne-6\)
⇔ \(\frac{720\left(x+6\right)}{6x\left(x+6\right)}=\frac{6x\left(x+6\right)}{6x\left(x+6\right)}+\frac{x\left(x+6\right)}{6x\left(x+6\right)}+\frac{6x\left(120-x\right)}{6x\left(x+6\right)}\)
\(\Rightarrow720x+4320=6x^2+36x+x^2+6x+720x-6x^2\)
\(\Leftrightarrow6x^2+36x+x^2+6x+720x-6x^2-720x-4320=0\)
\(\Leftrightarrow x^2+42x-4320=0\)
\(\Leftrightarrow x^2+90x-48x-4320=0\)
\(\Leftrightarrow\left(x+90\right)\left(x-48\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+90=0\\x-48=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-90\\x=48\end{matrix}\right.\) ( tm )
4 tháng 3 2017
a) \(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2+x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\\left(x+\frac{1}{2}\right)^2=-\frac{3}{4}\end{cases}}}\Rightarrow\)Vô lí
b)\(\Leftrightarrow\left(\frac{x+106}{3}-2\right)+\left(\frac{x+116}{4}-4\right)+\left(\frac{x+130}{5}-6\right)+\left(\frac{x-148}{6}-8\right)=0\Leftrightarrow\frac{x+100}{3}+\frac{x+100}{4}+\frac{x+100}{5}+\frac{x+100}{6}=0\Leftrightarrow\left(x+100\right)\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)=0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
17 tháng 3 2019
\(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne30\\x\ne24\end{cases}}\)
Ta có \(\frac{60}{\frac{120}{x}-4}+\frac{60}{\frac{120}{x}-5}=x\)
\(\Leftrightarrow\frac{60}{\frac{120-4x}{x}}+\frac{60}{\frac{120-5x}{x}}=x\)
\(\Leftrightarrow\frac{60x}{120-4x}+\frac{60x}{120-5x}=x\)
\(\Leftrightarrow\frac{60}{120-4x}+\frac{60}{120-5x}=1\left(Do\text{ }x\ne0\right)\)
\(\Leftrightarrow\frac{15}{30-x}=1-\frac{12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{24-x-12}{24-x}\)
\(\Leftrightarrow\frac{15}{30-x}=\frac{12-x}{24-x}\)
\(\Leftrightarrow360-15x=\left(12-x\right)\left(30-x\right)\)
\(\Leftrightarrow360-15x=360-42x+x^2\)
\(\Leftrightarrow x^2-27x=0\)
\(\Leftrightarrow x\left(x-27\right)=0\)
\(\Leftrightarrow x=27\left(Tm\text{ }ĐKXĐ\right)\)
4 tháng 3 2017
b. \(\dfrac{x+106}{3}+\dfrac{x+116}{4}+\dfrac{x+130}{5}+\dfrac{x+148}{6}=0\)\(\Leftrightarrow\dfrac{x+106}{3}+\dfrac{x+116}{4}+\dfrac{x+130}{5}+\dfrac{x+148}{6}-20=0\)\(\Leftrightarrow\dfrac{x+106}{3}-2+\dfrac{x+116}{4}-4+\dfrac{x+130}{5}-6+\dfrac{x+148}{6}-8=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}\ne0\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy PT có nghiệm \(x=-100\)
5 tháng 3 2017
\(x^4+x^3+2x^2+x+1=0\\ \Leftrightarrow\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)=0\\ \Leftrightarrow x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x^2+x+1\right)\left(x^2+1\right)=0\\ \)
Vì x^2+x+1\(>0\) với mọi x và x^2+1\(>0\) với mọi x nên (x^2+x+1)(x^2+1)>0 với mọi x
Vậy phương trình vô nghiệm
20 tháng 1 2018
ta có x2+5x+4
=x2+x+4x+4
=(x2+x)+(4x+4)
=x(x+1)+4(x+1)
=(x+1)(x+4)
tương tự ta đc
x2+11x+28=(x+4)(x+7)
x2+17x+70=(x+7)(x+10)
x2+23x+130=(x+10)(x+13)
=>\(\dfrac{1}{\left(x+1\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+7\right)}+\dfrac{1}{\left(x+7\right)\left(x+10\right)}+\dfrac{1}{\left(x+10\right)\left(x+13\right)}=\dfrac{4}{13}\)\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{3}{\left(x+4\right)\left(x+7\right)}+\dfrac{3}{\left(x+7\right)\left(x+10\right)}+\dfrac{3}{\left(x+10\right)\left(x+11\right)}=\dfrac{4}{13}\)=>\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}+....+\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{1}{x+1}-\dfrac{1}{x+13}=\dfrac{4}{13}\)
=>\(\dfrac{13\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}-\dfrac{13\left(x+1\right)}{13\left(x+1\right)\left(x+13\right)}=\dfrac{4\left(x+1\right)\left(x+13\right)}{13\left(x+1\right)\left(x+13\right)}\)
=> 13(x+13)-13(x+1)=4(x+1)(x+13)
=> 13[(x+13)-(x+1)]=(4x+4)(x+13)
=>13(x+13-x-1)=4x2+52x+4x+52
=13.12=4x2+56x+52
=>4x2+56x+52=156
=>4x2+56x-104=0
\(x\cdot120\%+\left(350-x\right)\cdot130\%=435\)
=>\(1,2x+455-1,3x=435\)
=>-0,1x=-20
=>x=200