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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: \(7^6+7^5-7^4⋮11=7^4\left(7^2+7-1\right)=7^4.55⋮11\)
\(\Rightarrowđpcm\)
1.
a) \(\frac{3}{7}+\frac{-5}{2}+\frac{-3}{5}\\ =\frac{30}{70}+\frac{-175}{70}+\frac{-42}{70}\\ =\frac{30-175-42}{70}\\ =\frac{-187}{70}\)
b) \(\frac{-8}{18}-\frac{15}{27}\\ =\frac{-4}{9}-\frac{5}{9}\\ =\frac{-9}{9}=-1\)
c) \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\\ =\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\\ =\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\\ =\frac{56+20-49}{70}\\ =\frac{27}{70}\)
2.
a) \(x+\frac{1}{4}=\frac{4}{3}\\ x=\frac{4}{3}-\frac{1}{4}\\ x=\frac{16}{12}-\frac{3}{12}\\ x=\frac{13}{12}\)
Vậy \(x=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=\frac{-6}{7}\\ -x=\frac{-6}{7}+\frac{2}{3}\\ -x=\frac{-18}{21}+\frac{14}{21}\\ -x=-\frac{4}{21}\\ x=\frac{4}{21}\)
Vậy \(x=\frac{4}{21}\)
c) \(x^2=16\\ x^2=4^2=\left(-4\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{4;-4\right\}\)
d) \(\frac{2}{3}+\frac{5}{3}x=\frac{5}{7}\\ \frac{5}{3}x=\frac{5}{7}-\frac{2}{3}\\ \frac{5}{3}x=\frac{15}{21}-\frac{14}{21}\\ \frac{5}{3}x=\frac{1}{21}\\ x=\frac{1}{21}:\frac{5}{3}\\ x=\frac{1}{21}\cdot\frac{3}{5}\\ x=\frac{1}{35}\)
Vậy \(x=\frac{1}{35}\)
3.
a) Xét △AKB và △AKC có:
AB = AC
KB = KC
AK: cạnh chung
\(\Rightarrow\text{△AKB = △AKC (c.c.c) }\)
b) \(\text{△AKB = △AKC }\)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà \(\widehat{AKB}+\widehat{AKC}=180^o\) (2 góc kề bù)
\(\Rightarrow\widehat{AKB}=\widehat{AKC}=90^o\\ \Rightarrow AK\perp BC\)
Câu 3:
a/ Xét ΔAKB và ΔAKC có:
AB = AC (GT)
\(\widehat{BAK}=\widehat{CAK}\left(GT\right)\)
AK: cạnh chung
=> ΔAKB = ΔAKC (c.g.c)
b/ VìΔAKB = ΔAKC (câu a)
\(\widehat{AKB}=\widehat{AKC}\) (2 góc tương ứng)
Mà 2 góc này lại là hai góc kề bù
=> \(\widehat{AKB}=\widehat{AKC}=180^0:2=90^0\)
=> AK ⊥BC
Cau 2:
a) \(x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
b) \(-x-\frac{2}{3}=-\frac{6}{7}\)
=> \(-x=-\frac{6}{7}+\frac{2}{3}=-\frac{4}{21}\)
=> \(x=\frac{4}{21}\)
c) x2 = 16
=> x = 4 hoặc x =-4
c: \(=\dfrac{7}{23}\cdot\left(\dfrac{-4}{3}-\dfrac{5}{2}\right)=\dfrac{7}{23}\cdot\dfrac{-8-15}{6}\)
\(=\dfrac{7}{23}\cdot\dfrac{-23}{6}=-\dfrac{7}{6}\)
d: \(=\dfrac{5}{7}\left(23+\dfrac{1}{4}-13-\dfrac{1}{4}\right)=\dfrac{5}{7}\cdot10=\dfrac{50}{7}\)
e: \(=\dfrac{2^5\cdot3^3\cdot5^3}{2^3\cdot3^3\cdot2^2\cdot5^2}=5\)
i: \(=\dfrac{1}{3^{10}}\cdot3^{50}-\dfrac{2^{10}}{3^{10}}:\dfrac{4^5}{3^{10}}\)
\(=3^{40}-1\)
a) \(2\left(4x-30\right)-3\left(x+5\right)+4\left(x-10\right)=5\left(x+2\right)\)
\(\Leftrightarrow8x-60-3x+15+4x-40=5x+10\)
\(\Leftrightarrow9x-35=5x+10\)
\(\Leftrightarrow9x-5x=10+35\)
\(\Leftrightarrow4x=45\)
\(\Leftrightarrow x=\dfrac{45}{4}=11,25\)
b) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\left(6x+1\right)\)
\(\Leftrightarrow\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}+x=4x+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{31}{60}-\dfrac{2}{3}=4x-x\)
\(\Leftrightarrow3x=\dfrac{1}{60}\)
\(\Leftrightarrow x=\dfrac{1}{180}\)
c) \(\dfrac{7}{3}-\left(2x-\dfrac{1}{3}\right)=\left(-2\dfrac{1}{6}+1\dfrac{1}{2}\right):0,25\)
\(\Leftrightarrow\dfrac{7}{3}-2x+\dfrac{1}{3}=-1\dfrac{2}{3}:\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-5}{3}.4\)
\(\Leftrightarrow\dfrac{8}{3}-2x=\dfrac{-20}{3}\)
\(\Leftrightarrow2x=\dfrac{8}{3}+\dfrac{20}{3}\)
\(\Leftrightarrow2x=\dfrac{28}{3}\)
\(\Leftrightarrow x=4\dfrac{2}{3}\)
d) \(0,75+\dfrac{5}{9}:x=5\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{5}{9}:x=\dfrac{11}{2}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{11}{2}-\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{5}{9}:x=\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{5}{9}:\dfrac{19}{4}\)
\(\Leftrightarrow x=\dfrac{20}{171}\)
a) \(x+\dfrac{1}{3}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{12}\) vậy \(x=\dfrac{5}{12}\)
b) \(x-\dfrac{2}{5}=\dfrac{5}{7}\Leftrightarrow x=\dfrac{5}{7}+\dfrac{2}{5}\Leftrightarrow x=\dfrac{39}{35}\) vậy \(x=\dfrac{39}{35}\)
c) \(-x-\dfrac{2}{3}=\dfrac{-6}{7}\Leftrightarrow x=\dfrac{-2}{3}+\dfrac{6}{7}\Leftrightarrow x=\dfrac{4}{21}\) vậy \(x=\dfrac{4}{21}\)
d) \(\dfrac{4}{7}-x=\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{7}-\dfrac{1}{3}\Leftrightarrow x=\dfrac{5}{21}\) vậy \(x=\dfrac{5}{21}\)
a) x + \(\dfrac{1}{3}\) = \(\dfrac{3}{4}\)
x = \(\dfrac{3}{4}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{5}{12}\)
Vậy x = \(\dfrac{5}{12}\)
b) x - \(\dfrac{2}{5}\) = \(\dfrac{5}{7}\)
x = \(\dfrac{5}{7}\) + \(\dfrac{2}{5}\)
x = \(\dfrac{39}{35}\)
Vậy x = \(\dfrac{39}{35}\)
c) -x - \(\dfrac{2}{3}\) = \(-\dfrac{6}{7}\)
- x = \(-\dfrac{6}{7}\) + \(\dfrac{2}{3}\)
- x = \(-\dfrac{4}{21}\)
⇒ x = \(\dfrac{4}{21}\)
Vậy x = \(\dfrac{4}{21}\)
d) \(\dfrac{4}{7}\) - x = \(\dfrac{1}{3}\)
x = \(\dfrac{4}{7}\) - \(\dfrac{1}{3}\)
x = \(\dfrac{5}{21}\)
Vậy x = \(\dfrac{5}{21}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
\(\left(\frac{1}{2}\right)^5\times x=\left(\frac{1}{2}\right)^7\)
\(x=\left(\frac{1}{2}\right)^7\div\left(\frac{1}{2}\right)^5\)
\(x=\left(\frac{1}{2}\right)^{7-5}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\) .
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{9}{21}\right)^2\)
\(\left(\frac{3}{7}\right)^2\times x=\left(\frac{3}{7}\right)^4\)
\(x=\left(\frac{3}{7}\right)^4\div\left(\frac{3}{7}\right)^2\)
\(x=\left(\frac{3}{7}\right)^{4-2}=\left(\frac{3}{7}\right)^2=\frac{9}{49}\)
\(2^x=2\Rightarrow x=1\)
\(3^x=3^4\Rightarrow x=4\)
\(7^x=7^7\Rightarrow x=7\)
\(\left(-3\right)^x=\left(-3\right)^5\Rightarrow x=5\)
\(\left(-5\right)^x=\left(-5\right)^4\Rightarrow x=4\)
\(2^x=4\Leftrightarrow2^x=2^2\Rightarrow x=2\)
\(2^x=8\Leftrightarrow2^x=2^3\Rightarrow x=3\)
\(2^x=16\Leftrightarrow2^x=2^4\Rightarrow x=4\)
\(3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=2-1\Rightarrow x=1\)
\(5^{x-1}=5\Leftrightarrow x-1=1\Leftrightarrow x=1+1\Rightarrow x=2\)
\(6^{x+4}=6^{10}\Leftrightarrow x+4=10\Leftrightarrow x=10-4\Rightarrow x=6\)
\(5^{2x-7}=5^{11}\Leftrightarrow2x-7=11\Leftrightarrow2x=11+7\Leftrightarrow2x=18\Leftrightarrow x=18\div2\Rightarrow x=9\)
\(\left(-2\right)^{4x+2}=64\)
\(2^{-4x+2}=2^6\Leftrightarrow-4x+2=6\Leftrightarrow-4x=6-2\Leftrightarrow-4x=4\Leftrightarrow x=4\div\left(-4\right)\Rightarrow x=-1\)
\(\left(\frac{1}{2}\right)^x=\left(\frac{1}{2}\right)^5\Rightarrow x=5\)
\(\left(\frac{5}{6}\right)^{2x}=\left(\frac{5}{6}\right)^5\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\Rightarrow2x-1=5x-4\)
\(2x-5x=-4+1\)
\(-3x=-3\Rightarrow x=1\)
\(\left(\frac{-1}{10}\right)^x=\frac{1}{100}\)
\(\left(\frac{1}{10}\right)^{-x}=\left(\frac{1}{10}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{2}\right)^x=\frac{9}{4}\)
\(\left(\frac{3}{2}\right)^{-x}=\left(\frac{3}{2}\right)^2\Rightarrow-x=2\Rightarrow x=-2\)
\(\left(\frac{-3}{5}\right)^{2x}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{-2x}=\left(\frac{3}{5}\right)^2\Rightarrow-2x=2\Rightarrow x=-1\)
\(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\Rightarrow x=3\).
hehe.
đánh tới què tay, hoa mắt lun r nekkk!!
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 \(3^{x+3}+3^{x+1}+2^{x+3}+2^{x+2}⋮6\)