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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
bài 1 : thực hiện phép tính
a) 3.52+15.22-26:2
= 3.25 + 15.4 - 26 : 2
= 75 + 60 - 13
= 135 - 13
= 122
b) 20:22+59:58
= 20:4 + 5
= 5 + 5
= 10
c) 100:52+7.32
= 100:25 + 7.9
= 4 + 63
= 67
d) 295-(31-22.5)2
= 295-(31-4.5)2
= 295 - 112
= 295 - 121
= 174
e) (-47)-[(45.24-52.12):14]
= (-47)-[(45.16-25.12):14]
= (-47)-[(720-300):14]
= (-47)-( 420:14 )
= (-47) - 30
= -77
f) (-2011)+5.[300-(17-7)2]
= (-2011)+5.(300-102)
= (-2011)+5.(300-100)
= (-2011)+5.200
= (-2011)+1000
= -1011
g) 5.[29-(6-1)2]-129
= 5.(29-52)-129
= 5.(29-25)-129
= 5.4-129
= 20-129
= -109
Đúng thì tik cái nha ! Thanks nhiều ! 
Bài giải
a) Ta có :
\(43^{43}-17^{17}=43^{40}\cdot43^3-17^{16}\cdot17=\left(43^4\right)^{10}\cdot43^3-\left(17^4\right)^4\cdot17=\overline{\left(...1\right)}^{10}\cdot\overline{\left(...3\right)}^3-\overline{\left(...1\right)}^4\cdot17\)
\(=\overline{\left(...1\right)}\cdot\overline{\left(...7\right)}-\overline{\left(...7\right)}=\overline{\left(...7\right)}-\overline{\left(...7\right)}=\overline{\left(...0\right)}\text{ }⋮\text{ }10\)
\(\Rightarrow\text{ ĐPCM}\)
Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)
=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)
=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)
=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)
=> \(x\) = \(\dfrac{2}{5}\)
4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)
=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)
=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)
=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)
=> \(3x=\dfrac{1}{9}\)
=> \(x=\dfrac{1}{9}:3\)
=> \(x=\dfrac{1}{27}\)
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