1/4+2.(3.x-2/3)=1 ai giúp mình làm ik đc ko
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\(\frac{3}{8}\times\frac{1}{6}+\frac{1}{6}\times\frac{5}{8}\)
\(=\frac{1}{6}\times\left(\frac{3}{8}+\frac{5}{8}\right)\)
\(=\frac{1}{6}\times1=\frac{1}{6}\)
Học kĩ cộng trừ nhân chia phân số nhiều vào
A = 1 . 3 + 3 . 5 + 5 . 7 + ... + 49 . 51
A=1*51
A=
B = 2 . 4 + 4 . 6 + 6 . 8 + ... + 98 . 100
B=2*100
B=200
C = 1 . 4 + 4 . 7 + 7 . 10 + ... + 301 . 304
C=1*304
C=304
D = 1 + 1 . 1! + 2 . 2! + 3 . 3! + ... + 100 . 100!
D=1*100!
D=100!
E = 22 + 42 + ... + ( 2n )2
E=\(2^2\cdot2n^2\)
E=\(2n^4\)
a) <=> 16(2x-3) = 21(x+1) <=> 32x - 48 = 21x + 21 <=> 32x - 21x = 21+ 48 <=> 11x = 69 <=> x = 69 / 11
b) <=> 7(x-3) = 6(x-5) <=> 7x - 21 = 6x - 30 <=> 7x - 6x = 21 - 30 <=> x = -9
c) <=> 2x+3 = 6(7x-3) <=> 2x+3 = 42x - 18 <=>42x-2x = 18 + 3 <=>40x = 21 <=> x = 21/40
d) <=> -3(2-x) = 4(3x-1) <=> 3x - 6 = 12x - 4 <=> 12x - 3x = 4-6 <=> 9x = -2 <=> x=-2/9
phương pháp ở đây lafnhaan chéo hai vế nếu bạn thấy mk tính sai chỗ nào thì bạn tính lại nha nếu ko đc thì nhắn tin cho mk
Viết lại đề bài:
Tìm số nguyên x sao cho \(\frac{6}{x+1}.\frac{x-1}{3}\)là số nguyên
Giải:
\(\frac{6}{x+1}.\frac{x-1}{3}\text{}\)
\(=\frac{3.2}{x+1}.\frac{x-1}{3}\text{}\)
\(=\frac{3.2.\left(x-1\right)}{\left(x+1\right).3}\text{}\)
\(=\frac{3.2.\left(x-1\right)}{3.\left(x+1\right)}\)
\(=\frac{3.2.\left(x-1\right)}{3.\left(x+1\right)}\)
\(=\frac{2.\left(x-1\right)}{\left(x+1\right)}\)
\(=2.\frac{\left(x-1\right)}{\left(x+1\right)}\)
Bí....
Sorr nhak
Ta có:\(\frac{6x}{x+1}=\frac{6x+6-6}{x+1}=\frac{6\left(x+1\right)-6}{x+1}=6-\frac{6}{x+1}\)
Để\(\frac{6x}{x+1}\)là số nguyên \(\Leftrightarrow6⋮x+1\)
\(\Rightarrow x+1\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
\(\Rightarrow x=\left\{-7;-4;-3;-2;0;1;2;5\right\}\left(1\right)\)
Để\(\frac{x-1}{3}\)là số nguyên\(\Leftrightarrow\left(x-1\right)⋮3\)
\(\Rightarrow x-1=3k\Rightarrow x=3k+1\left(k\in Z\right)\left(2\right)\)
Từ (1) và (2)\(\Rightarrow x\in\left\{-2;1\right\}\)
Vậy \(x\in\left\{-2;1\right\}\)
= (x^2 -1).(x+2)-(x-2).(x+2)^2
=(x^2-1).(x+2)-(x-2).(x+2).(x+2)
=(x^2-1).(x+2)-(x^2-2^2)(x+2)
=(x^2-1-x^2+4)(x+2)
a, 7\(\dfrac{3}{5}\) : \(x\) = 5\(\dfrac{4}{15}\) - 1\(\dfrac{1}{6}\)
\(\dfrac{38}{5}\) : \(x\) = \(\dfrac{79}{15}\) - \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{41}{10}\)
\(x\) = \(\dfrac{38}{5}\) : \(\dfrac{41}{10}\)
\(x\) = \(\dfrac{76}{41}\)
b, \(x\) \(\times\) 2\(\dfrac{2}{3}\) = 3\(\dfrac{4}{8}\) + 6\(\dfrac{5}{12}\)
\(x\) \(\times\) \(\dfrac{8}{3}\) = \(\dfrac{7}{2}\) + \(\dfrac{77}{12}\)
\(x\) \(\times\) \(\dfrac{8}{3}\) = \(\dfrac{119}{12}\)
\(x\) = \(\dfrac{119}{12}\)
\(x\) = \(\dfrac{119}{12}\): \(\dfrac{8}{3}\)
\(x\) = \(\dfrac{119}{32}\)
\(\dfrac{1}{4}\) + 2.(3\(x\) - \(\dfrac{2}{3}\)) = 1
2.(3\(x\) - \(\dfrac{2}{3}\)) = 1 - \(\dfrac{1}{4}\)
2.(3\(x\) - \(\dfrac{2}{3}\)) = \(\dfrac{3}{4}\)
3\(x\) - \(\dfrac{2}{3}\) = \(\dfrac{3}{4}\): 2
3\(x\) = \(\dfrac{3}{8}\) + \(\dfrac{2}{3}\)
3\(x\) = \(\dfrac{25}{24}\)
\(x\) = \(\dfrac{25}{24}\) : 3
\(x\) = \(\dfrac{25}{72}\)
Vậy \(x\) = \(\dfrac{25}{72}\)