1.tính bằng hai cách
(2/3-4/5)x25 3/4x2/503/4x2/7
2.tìm x
x:2/5=2/3 x:2/3=4/9
K
Khách
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HK
2
NK
19 tháng 3 2023
\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\left(\dfrac{4}{7}+5\right):\dfrac{4}{3}=\dfrac{39}{7}:\dfrac{4}{3}=\dfrac{39}{7}\times\dfrac{3}{4}=\dfrac{117}{28}\)
\(\dfrac{4}{7}:\dfrac{4}{3}+5:\dfrac{4}{3}=\dfrac{3}{7}+\dfrac{15}{4}=\dfrac{12}{28}+\dfrac{105}{28}=\dfrac{117}{28}\)
\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{4}\times\left(\dfrac{2}{5}-\dfrac{2}{7}\right)=\dfrac{3}{4}\times\dfrac{4}{35}=\dfrac{12}{140}=\dfrac{3}{35}\)
\(\dfrac{3}{4}\times\dfrac{2}{5}-\dfrac{3}{4}\times\dfrac{2}{7}=\dfrac{3}{10}-\dfrac{3}{14}=\dfrac{42}{140}-\dfrac{30}{140}=\dfrac{12}{140}=\dfrac{3}{35}\)
\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{21}\times\dfrac{2}{5}=\dfrac{4}{105}\)
\(\left(\dfrac{2}{3}-\dfrac{4}{7}\right)\times\dfrac{2}{5}=\dfrac{2}{3}\times\dfrac{2}{5}-\dfrac{4}{7}\times\dfrac{2}{5}=\dfrac{4}{15}-\dfrac{8}{35}=\dfrac{140}{525}-\dfrac{120}{525}=\dfrac{20}{525}=\dfrac{4}{105}\)
#YVA
19 tháng 3 2023
4/7:4/3+5:4/3
= 3/7 + 15/4
=4 5/28
4/7:4/3+5:4/3
= (4/7+5) x 3/4
=4 5/28
3/4x2/5-3/4x2/7
=3/10 - 3/14
=3/35
3/4x2/5-3/4x2/7
=3/4x(2/5-2/7)
=3/35
(2/3-4/7)x2/5
=2/21x2/5
=4/105
(2/3-4/7)x2/5
=2/3x2/5-4/7x2/5
=4/105
NH
2
S
9 tháng 1 2017
a) C1 : 2/5 x 3/7 + 2/7 x 4/7
= 2/5 x (3/7 + 4/7)
= 2/5 x 1
= 2/5
C2 : 2/5 x 3/7 + 2/5 x 4/7
= 6/35 + 8/35
= 2/5
b) (2/3 - 4/7) x 5/5
= 2/21 x 5/5
= 2/21
C2 : (2/3 - 4/7) x 5/5
= 2/3 x 1 - 4/7 x 1
= 2/3 - 4/7
= 2/21
c) 3/4 x 2/5 - 3/4 x 2/7
= 3/4 x (2/5 - 2/7)
= 3/4 x 4/35
= 3/35
C2 : 3/4 x 2/5 - 3/4 x 2/7
= 3/10 - 3/14
= 3/35
9 tháng 1 2017
a,
Cách 1
\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)
\(=\frac{6}{35}+\frac{8}{35}\)
\(=\frac{2}{5}\)
Cách 2 :
\(\frac{2}{5}\times\frac{3}{7}+\frac{2}{5}\times\frac{4}{7}\)
\(=\frac{2}{5}\times\left(\frac{3}{7}+\frac{4}{7}\right)\)
\(=\frac{2}{5}\times1\)
\(=\frac{2}{5}\)
b,
Cách 1 :
\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)
\(=\frac{2}{21}\times1\)
\(=\frac{2}{21}\)
Cách 2 :
\(\left(\frac{2}{3}-\frac{4}{7}\right)\times\frac{5}{5}\)
\(=\frac{2}{3}\times\frac{5}{5}-\frac{4}{7}\times\frac{5}{5}\)
\(=\frac{2}{3}-\frac{4}{7}\)
\(=\frac{2}{21}\)
c,
Cách 1 :
\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)
\(=\frac{3}{10}-\frac{3}{14}\)
\(=\frac{3}{35}\)
Cách 2 :
\(\frac{3}{4}\times\frac{2}{5}-\frac{3}{4}\times\frac{2}{7}\)
\(=\frac{3}{4}\times\left(\frac{2}{5}-\frac{2}{7}\right)\)
\(=\frac{3}{4}\times\frac{4}{35}\)
\(=\frac{3}{35}\)
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Đúng 100%
Tk mk mk tk lại
Cảm ơn bạn nhiều
Thank you very much
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17 tháng 9 2021
1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
18 tháng 9 2021
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
LP
1
14 tháng 8 2021
a: Ta có: \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x-5\right)=-4\)
\(\Leftrightarrow x^2+5x+6-x^2+7x-10=-4\)
\(\Leftrightarrow12x=0\)
hay x=0
b: Ta có: \(\left(x+1\right)\left(x^2-x+1\right)-x\left(x-3\right)\left(x+3\right)=8\)
\(\Leftrightarrow x^3+1-x^3+9x=8\)
\(\Leftrightarrow9x=7\)
hay \(x=\dfrac{7}{9}\)
c: Ta có: \(4x^2-9=\left(3x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x+1-2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\)
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