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a) 3x+3x+2=812
Suy ra 3x+3x.32=812
3x.(1+32) =812
3x.10 =812
3x =812:10
3x =406/5
Suy ra x ko có giá trị
b)4\(\frac{1}{3}\):\(\frac{x}{4}\)=6:0,3
suy ra \(\frac{13}{3}\):\(\frac{x}{4}\) =20
x/4 = 13/3:20
x/4 = 13/60
x = 13/15
c) I 2x + 0,5I=8,5
2x+0,5=8,5 hoặc 2x+0,5=-8,5
TH1:2x+0,5=8,5=>x=4
TH2:2x+0,5=-8,5=>x=-9/2
d) 8x: 2x =1635
=>(8:2)x=1635
=>4x =1635
=>4x =(42)35
=>4x =42.35 =>4x=470 =>x=70
Vậy x = 70
Đầy đủ và chính xác lắm đó.
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}.\)
\(\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1.\)(cộng 2 vế cho 3)
\(\frac{x+1}{2009}+\frac{2009}{2009}+\frac{x+2}{2008}+\frac{2008}{2008}+\frac{x+3}{2007}+\frac{2007}{2007}=\frac{x+10}{2000}+\frac{2000}{2000}+\frac{x+11}{1999}+\frac{1999}{1999}+\frac{x+12}{1998}+\frac{1998}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}.\)
\(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
x+2010=0
x=-2010
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Leftrightarrow\left(1+\frac{x+1}{2009}\right)+\left(1+\frac{x+2}{2008}\right)+\left(1+\frac{x+3}{2007}\right)\)
\(=\left(1+\frac{x+10}{2000}\right)+\left(1+\frac{x+11}{1999}\right)+\left(1+\frac{x+12}{1998}\right)\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x=2010}{1998}\)
\(\Leftrightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}\)
\(=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
\(\Leftrightarrow x+2010=0\)
\(\Leftrightarrow x=-2010\)
\(\frac{1}{3}x-\frac{3}{5}=\frac{5}{6}x+2\)
\(\Leftrightarrow\frac{x}{3}-\frac{3}{5}=\frac{5x}{6}+2\)
\(\Leftrightarrow2x-\frac{18}{5}=5x+12\)
\(\Leftrightarrow2x-5x=\frac{18}{5}+12\)
\(\Leftrightarrow-3x=\frac{78}{5}\)
\(\Leftrightarrow3x=-\frac{78}{5}\)
\(\Leftrightarrow x=-\frac{26}{5}\)
Ps: đoạn nào không hiểu hỏi anh nhé. Nhớ k để tạo động lực cho anh nhé :33
# Aeri #
\(m\left(x\right)+h\left(x\right)=g\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=g\left(x\right)-h\left(x\right)-5\)
\(\Leftrightarrow m\left(x\right)=4x^2+3x+1-3x^2+2x+3-5\)
\(\Leftrightarrow m\left(x\right)=x^2+5x-1\)
\(\left(\frac{2}{7}\right)^{2008}:\left(\frac{4}{49}\right)^{1004}\)
\(=\left(\frac{2}{7}\right)^{2008}:\left[\left(\frac{2}{7}\right)^2\right]^{1004}\)
\(=\left(\frac{2}{7}\right)^{2008}:\left(\frac{2}{7}\right)^{2008}\)
= 1
Học tốt
#Gấu
( \(\frac{x}{2}\)+ 3 )2 = 16
( \(\frac{x}{2}\)+ 3 )2 = 42
=> \(\frac{x}{2}\)+ 3 = 4
\(\frac{x}{2}\)= 4 - 3
\(\frac{x}{2}\)= 1
x = 1 x 2
x = 2
#Chúc em học tốt
\(\left(x^2+3\right)^2=16\)
=> \(\orbr{\begin{cases}\frac{x}{2}+3=4\\\frac{x}{2}+3=-4\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{x}{2}=1\\\frac{x}{2}=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-14\end{cases}}\)
Vậy \(x\in\left\{2;-14\right\}\)là giá trị cần tìm