Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
bài 2
a] = 3 x \(\frac{4343}{7171}\)= \(\frac{17372}{7171}\)= \(\frac{172}{71}\)
b] = \(\frac{1}{33}\)x \(\frac{44}{7}\)= \(\frac{1}{3}\)x \(\frac{4}{7}\)=\(\frac{4}{21}\)
bài 1
a] y là 9
b] <=> 64y + 36y = 700 - 75 - 225
<=> 100y = 400
<=> y = 4
trên lớp cô sửa rồi nên mình giải luôn:
1) Tìm y
a) y3 + 3y = 12 x 11
y3 + 3y = 132
y x 10 + 3 + 3 x 10 + y = 132
( y x 10 + y ) + ( 3 x 10 + 3 ) = 132
11 x y + 33 = 132
11 x y = 132 - 33
11 x y = 99
y = 99 : 11
y = 9
b) 64 x y + 225 = 700 - 75 - 36 x y
64 x y + 225 = 625 - 36 x y
64 x y + 36 x y = 625 -225
64 x y + 36 x y = 400
( 64 + 36 ) x y = 400
100 x y = 400
y = 400 : 100
y = 4
2) Tính
a) \(\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}+\frac{4343}{7171}\)
\(=\frac{4343}{7171}\times4\)
\(=\frac{43}{71}\times4\)
\(=\frac{172}{71}\)
b) A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
Ta có:
\(\frac{3333}{2020}=\frac{3333:101}{2020:101}=\frac{33}{20}\)
\(\frac{333333}{303030}=\frac{333333:10101}{303030:10101}=\frac{33}{30}\)
\(\frac{33333333}{42424242}=\frac{33333333:1010101}{42424242:1010101}=\frac{33}{42}\)
A = \(\frac{1}{33}\times\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
A = \(\frac{1}{33}\times33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
A = 1 x \(\left(\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{1}{3}-\frac{1}{7}\right)\)
A = 1 x \(\left(\frac{7}{21}-\frac{3}{21}\right)\)
A = 1 x \(\frac{4}{21}\)
A = \(\frac{4}{21}\)
\(1,\\ a,=\dfrac{7}{19}\times\left(\dfrac{1}{3}+\dfrac{2}{3}\right)=\dfrac{7}{19}\times1=\dfrac{7}{19}\\ b,=\dfrac{2}{5}+\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=\dfrac{2}{5}+1=\dfrac{7}{5}\\ 2,\\ a,=15\times\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)\\ =15\times\left(\dfrac{2121:101}{4343:101}+\dfrac{222222:10101}{434343:10101}\right)\\ =15\times\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\times\dfrac{43}{43}=15\times1=15\)
\(3,\)
Cạnh \(AC=\) chu vi ABC \(-AB-BC=\dfrac{4}{5}-\dfrac{1}{5}-\dfrac{1}{4}=\dfrac{3}{5}-\dfrac{1}{4}=\dfrac{7}{20}\left(m\right)\)
Vì \(\dfrac{7}{20}>\dfrac{5}{20}>\dfrac{4}{20}\Rightarrow\dfrac{7}{20}>\dfrac{1}{4}>\dfrac{1}{5}\) nên \(AC>BC>AB\)
\(=15x\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15x1=15\)
\(a,=15\left(\dfrac{2121}{4343}+\dfrac{222222}{434343}\right)=15\left(\dfrac{21}{43}+\dfrac{22}{43}\right)=15\cdot1=15\)
Bài 1
a; \(\dfrac{7}{19}\) x \(\dfrac{1}{3}\) + \(\dfrac{7}{19}\) x \(\dfrac{2}{3}\)
= \(\dfrac{7}{19}\) x (\(\dfrac{1}{3}+\dfrac{2}{3}\))
= \(\dfrac{7}{19}\) x 1
= \(\dfrac{7}{19}\)
b; 15 x \(\dfrac{2121}{4343}\) + 15 x \(\dfrac{212121}{434343}\)
= 15 x \(\dfrac{21}{43}\) + 15 x \(\dfrac{21}{43}\)
= 15 x \(\dfrac{21}{43}\) x (1 + 1)
= 15 x \(\dfrac{21}{43}\) x 2
= (15 x 2) x \(\dfrac{21}{43}\)
= 30 x \(\dfrac{21}{43}\)
= \(\dfrac{630}{43}\)
a, \(\dfrac{15\times14-1}{13\times15+14}\)
= \(\dfrac{15\times\left(13+1\right)-1}{13\times15+14}\)
= \(\dfrac{13\times15+15-1}{13\times15+14}\)
= \(\dfrac{13\times15+14}{13\times15+14}\)
= 1
c, \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)
= \(\dfrac{\left(9+1\right)\times\left\{\left(9-1\right):1+1\right\}:2}{10}\)
= \(\dfrac{10\times9:2}{10}\)
= \(\dfrac{9}{2}\)