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Ta có: \(\frac{2}{3}a=\frac{1}{4}b\)
\(\Leftrightarrow\frac{2a}{3}=\frac{b}{4}\)
\(\Leftrightarrow2a=\frac{3b}{4}\)
hay \(a=\frac{3b}{4}:2=\frac{3b}{8}\)
Ta có: \(\frac{1}{2}b=\frac{1}{3}c\)
\(\Leftrightarrow\frac{b}{2}=\frac{c}{3}\)
hay \(c=\frac{3b}{2}\)
Ta có: a+b+c=90
\(\Leftrightarrow\frac{3b}{8}+b+\frac{3b}{2}=90\)
\(\Leftrightarrow b\left(\frac{3}{8}+1+\frac{3}{2}\right)=90\)
\(\Leftrightarrow b\cdot\frac{23}{8}=90\)
hay \(b=90:\frac{23}{8}=\frac{720}{23}\)
Ta có: \(a=\frac{3b}{8}\)(cmt)
hay \(a=3\cdot\frac{720}{23}:8=\frac{270}{23}\)
Ta có: a+b+c=90
\(\Leftrightarrow c=90-a-b=90-\frac{270}{23}-\frac{720}{23}=\frac{1080}{23}\)
Vậy: \(\left(a,b,c\right)=\left(\frac{270}{23};\frac{720}{23};\frac{1080}{23}\right)\)
Giải:
Ta có:
\(\left(a+b+c+d\right)-\left(a+c+d\right)._{\left(1\right)}\)
\(=a+b+c+d-a-c-d.\)
\(=\left(a-a\right)+\left(c-c\right)+\left(d-d\right)+b.\)
\(=0+0+0+b=b.\)
Thay số vào \(_{\left(1\right)}\)\(\Rightarrow1-2=b\Rightarrow b=-1\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+d\right)._{\left(2\right)}\)
\(=a+b+c+d-a-b-d.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(d+d\right)+c.\)
\(=0+0+0+c=c.\)
Thay số vào \(_{\left(2\right)}\)\(\Rightarrow1-3=c\Rightarrow c=-2\in Z.\)
\(\left(a+b+c+d\right)-\left(a+b+c\right)_{\left(3\right)}.\)
\(=a+b+c+d-a-b-c.\)
\(=\left(a-a\right)+\left(b-b\right)+\left(c-c\right)+d.\)
\(=0+0+0+d=d.\)
Thay số vào \(_{\left(3\right)}\)\(\Rightarrow1-4=d\Rightarrow d=-3\in Z.\)
\(\Rightarrow a+b+c+d=1.\)
\(a+\left(-1\right)+\left(-2\right)+\left(-3\right)=1.\)
\(\Rightarrow a=1-\left(-1\right)-\left(-2\right)-\left(-3\right).\)
\(\Rightarrow a=1+1+2+3=7\in Z.\)
Vậy \(\left\{a;b;c;d\right\}=\left\{7;-1;-2;-3\right\}.\)
Do a + b + c + d = 1 mà a + c + d = 2
=> b = 1 - 2 = -1
=> c = 1 - 3 = -2
=> d = 1 - 4 = -3
=> a = 1 - (-1 - 2 - 3) = 7
@Valentine
1. a, 3x + 2 \(⋮2x-1\)
Có 3(2x - 1) \(⋮2x-1\)
Và 2(3x - 2) \(⋮2x-1\)
=> 6x - 4 - 6x + 3 \(⋮2x-1\)
<=> -1 \(⋮2x-1\)
=> 2x - 1 \(\inƯ\left(1\right)=\left\{\pm1\right\}\)
=> 2x = 2; 0
=> x = 1; 0 (thỏa mãn)
@Lớp 6B Đoàn Kết
1. b, x2 - 2x + 3 \(⋮x-1\)
<=> x(x - 2) + 3 \(⋮x-1\)
<=> x(x - 1) - x + 3 \(⋮x-1\)
<=> x(x - 1) - (x - 1) - 2 \(⋮x-1\)
<=> (x - 1)2 - 2 \(⋮x-1\)
<=> -2 \(⋮x-1\)
=> x - 1 \(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> x = 2; 0; 3; -1 (thỏa mãn)
@Lớp 6B Đoàn Kết
ta có : abc = 100a + 10b + c (1)
cba = 100c + 10b + a = (n-2)2 (2)
lấy (2) trừ (1) ta có: 99(a - c) = 4n - 5 => 4n - 5 \(⋮\) 99
100 \(\le\) n2 - 1 \(\le\) 999
<=> \(101\le n^2\le1000\)
<=> \(11\le n\le31\)
<=> \(44\le4n\le124\)
<=> \(39\le4n-5\le119\)
mà 4n - 5 \(⋮\) 99
=> 4n - 5 = 99
=> n = 26
=>abc = 262 - 1 = 675
VẬy.....
a) 2017 + 5.[ 300 - \(\left(17-7\right)^2\)]
= 2017 + 5.[ 300 - \(10^2\)]
= 2017 + 5.[ 300 - 100]
= 2017 + 5. 200
= 2017 + 1000
= 3017
b) \(5^{27}\).5.\(5^{25}\)-|-125|
= \(5^{27}\). 5 . \(5^{25}\) - 125
= \(5^{53}\) - 125
= \(5^{53}\) - \(5^3\)
= \(5^{53}\)+ 3
c) (\(5^{25}\).18+ \(5^{15}\).7) : \(5^{17}\)
= [ (\(5^{25}\) . \(5^{15}\)) . ( 18 . 7) ] : \(5^{17}\)
= [ \(5^{40}\) . 126 ] : \(5^{17}\)
= [ \(5^{40}\) : \(5^{17}\) ] . 126
= \(5^{23}\) . 126
Phần c) chưa chắc làm đúng nha
Học tốt :'3