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a) ta có: \(\frac{a}{4}=\frac{b}{5};\frac{b}{5}=\frac{c}{8}\)
\(\Rightarrow\frac{a}{4}=\frac{b}{5}=\frac{c}{8}=\frac{5a}{20}=\frac{3b}{15}=\frac{3c}{24}\)
ADTCDTSBN
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bn tự áp dụng rùi tìm a;b;c nha
b) ta có: \(\frac{a+3}{5}=\frac{b-2}{3}=\frac{c-1}{7}=\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}\)
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có: \(\frac{3a+9}{15}=\frac{5b-10}{15}=\frac{7c-7}{49}=\frac{3a+9-5b+10+7c-7}{15-15+49}\)
\(=\frac{\left(3a-5b+7c\right)+\left(9+10-7\right)}{49}=\frac{86+12}{49}=\frac{98}{49}=2\)
=>...
c) ta cóL \(\frac{a}{7}=\frac{b}{6}\Rightarrow\frac{a}{35}=\frac{b}{30}\)
\(\frac{b}{5}=\frac{c}{8}\Rightarrow\frac{b}{30}=\frac{c}{48}\)
\(\Rightarrow\frac{a}{35}=\frac{b}{30}=\frac{c}{48}=\frac{2b}{60}\)
ADTCDTSBN
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các bài còn lại bn dựa vào mak lm nha!
a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
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, 7/12 - 3/4 . 5/6
= 7/12 - 5/8
= 14/24- 15/24
= -1/24
b,( 2/1/3 + 1/3/4 ) . 12/13
= ( 6/3 + 7/3 ) . 12/13
= 13/3 . 12/13
=4
c, 12 : ( 3/4 -5/6 ) . 2
= 12 : ( -1/12 ) .2
= 12 . -12 . 2
= -228
d, 7/22 : 3/11 + 7/22 : 4/11
= 7/22 . 11/3 + 7/22 . 11/4
= 7/22 . ( 11/3 + 11/4 )
....
tiếp theo bạn tự làm nhé!