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a, \(5^x+5^{x+1}+5^{x-2}=151\)
\(\Rightarrow5^x.\left(1+5+5^{-2}\right)=151\)
\(\Rightarrow5^x.6,04=151\Rightarrow5^x=25=5^2\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=2\)
b, \(5^{x-1}+5^{x-2}+5^{x-3}=155\)
\(\Rightarrow5^x.\left(5^{-1}+5^{-2}+5^{-3}\right)=155\)
\(\Rightarrow5^x.0,248=155\Rightarrow5^x=625=5^4\)
Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=4\)
c, \(5^{2+x}+5^{3+x}=750\) \(\Rightarrow5^x.\left(5^2+5^3\right)=750\) \(\Rightarrow5^x.150=750\Rightarrow5^x=5=5^1\) Vì \(5\ne-1;5\ne0;5\ne1\) nên \(x=1\) Chúc bạn học tốt!!!\(•5^x+5^{x+1}+5^{x-2}=151\\ 5^x\left(1+5+\dfrac{1}{25}\right)=151\\ 5^x=25\\ \Rightarrow x=2\)
\(•5^{x-1}+5^{x-2}+5^{x-3}=155\\ 5^x.\left(\dfrac{1}{5}+\dfrac{1}{25}+\dfrac{1}{125}\right)=155\\ 5^x=625\\ \Rightarrow x=4\)
\(•5^{2+x}+5^{3+x}=750\\ 5^x\left(25+125\right)=750\\ 5^x=5\\ \Rightarrow x=1\)
a) ta có : \(5^5-5^4+5^3=5^3.\left(5^2-5+1\right)=5^3.\left(25-5+1\right)\)
\(5^3.21=5^3.3.7⋮7\) (đpcm)
b) ta có : \(7^6+7^5-7^4=7^4.\left(7^2+7-1\right)=7^4.\left(49+7-1\right)\)
\(=7^4.55=7^4.5.11⋮11\) (đpcm)
c) ta có : \(3^{x+2}-2^{x+3}+3^x-2^{x+1}=3^{x+2}+3^x-2^{x+3}-2^{x+1}\)
\(=3^x\left(3^2+1\right)-2^x\left(2^3+2\right)=3^x.\left(9+1\right)-2^x.\left(8+2\right)\)
\(=3^x.10-2^x.10=10\left(3^x-2^x\right)⋮10\) (đpcm)
d) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}=3^x.\left(3^3+3\right)+2^x.\left(2^3+2^2\right)\)
\(=3^x.\left(27+3\right)+2^x\left(8+4\right)=3^x.30+2^x.12=6.\left(3^x.5+2^x.2\right)⋮6\) (đpcm)
a)Ta có:\(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21\)(vì 21 chia hết cho 7)
\(\)\(\RightarrowĐPCM\)
b)Ta có: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)
\(\Rightarrowđpcm\)
a) \(-12\left(x-5\right)+7\left(3-x\right)=15\)
\(\Leftrightarrow-12x+60+21-7x=15\)
\(\Leftrightarrow-19x=15-\left(60+21\right)=-66\)
\(\Leftrightarrow x=\frac{-66}{-19}=\frac{66}{19}\)
Vậy : \(x=\frac{66}{19}\)
b) \(-\left(2x-3\right)-4\left(x+1\right)=-7x+3\)
\(\Leftrightarrow-2x+3-4x-4=-7x+3\)
\(\Leftrightarrow-2x-4x+7x=3-3+4\)
\(\Leftrightarrow x=4\)
Vậy : \(x=4\)
c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}.\left(x-\frac{2}{3}\right)=\frac{3}{2}.\frac{-5}{6}\)
\(\Leftrightarrow-5x-1-\frac{1}{2}x+\frac{1}{3}=-\frac{5}{4}\)
\(\Leftrightarrow-\frac{11}{2}x=-\frac{5}{4}+1-\frac{1}{3}=-\frac{7}{12}\)
\(\Leftrightarrow x=-\frac{7}{12}:\frac{-11}{2}=\frac{7}{66}\)
Vậy : \(x=\frac{7}{66}\)
Câu d) Xíu làm mình bận >>
d) \(x-\left\{\left[-x+\left(x+3\right)\right]\right\}-\left[\left(x+3\right)-\left(x-2\right)\right]=0\)
\(\Leftrightarrow x-\left[-x+x+3\right]-\left[x+3-x+2\right]=0\)
\(\Leftrightarrow x+x-x-3-x-3+x-2=0\)
\(\Leftrightarrow\left(x+x-x-x+x\right)+\left(-3-3-2\right)=0\)
\(\Leftrightarrow x+\left(-8\right)=0\)
\(\Leftrightarrow x=8\)
Vậy : \(x=8\)
P/s : Câu này cần chú ý quy tắc chuyển dấu và quy tắc thực hiện khi có dấu ngoặc nhé !
Ta có \(5^5-5^4+5^3=5^3\left(5^2-5+1\right)=5^3.21=5^3.3.7\)
Vì 53.3 là số nguyên nên \(5^3.3.7⋮7\)
Vậy \(5^5-5^4+5^3⋮7\)
c) \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}\)
\(=\left(3^{x+3}+3^{x+1}\right)+\left(2^{x+3}+2^{x+2}\right)\)
\(=3^x\left(3^2+3\right)+2^x\left(2^2+2\right)\)
\(=3^x.12+2^x.6\)
\(=6\left(2.3^x+2^x\right)\)
Vì \(2.3^x+2^x\in Z\)
Nên : \(6\left(2.3^x+2^x\right)⋮6\)
Vậy \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)
Mk nhầm nha câu đầu chỉ có 1 cái x-1 + x -2 thôi ko có cái đằng sau nhé ! giá trị tuyệt đối thì vẫn giữ nguyên !