(x-2)2-(x+1)(x-3)=13
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HV
0
KC
2
H
10 tháng 7 2023
`1, -2/9 xx 15/17 + (-2/9) xx 2/17`
`= -2/9 xx (15/17 + 2/17)`
`= -2/9 xx 17/17`
`=-2/9xx1`
`=-2/9`
__
`-5/3 xx 6/5 + (-7/9) xx 3/10`
`= -30/15 + (-21/90)`
`= -2 + (-7/30)`
`=-60/30 +(-7/30)`
`=-67/30`
__
`15/20 xx 7/5 + (-9/7) xx (-6/4)`
`=3/4 xx7/5 + (-9/7) xx(-6/4)`
`= 21/20 + 54/28`
`= 21/20 + 27/14`
`=417/140`
__
`-25/13 xx 5/19 + (-25/13) xx 14/19`
`=-25/13 xx (5/19 +14/19)`
`=-25/13 xx 19/19`
`= -25/13 xx 1`
`=-25/13`
__
`-7/13 xx 13/5 + (-9/7) xx 5/3`
`=-7/5 +(-15/7)`
`=-124/35`
VH
16 tháng 6 2019
câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)
b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)
⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)
⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)
⇒ \(x=13\)
c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)
⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)
⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)
⇒ \(x=10\)
d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)
⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)
⇒ \(x=13\)
12 tháng 8 2020
11/13-(5/42-x)=(15/28-11/13)
11/13-(5/42-x)=-37/182
(5/42-x)=11/13+37/182
(5/42-x)=191/182
x=5/42-191/182
x=-254/273
vậy x=-254/273
10 tháng 7 2023
1: =-2/9(15/17+2/17)=-2/9
2: \(=\dfrac{-6}{3}+\dfrac{-21}{90}\)
=-2-7/30=-67/30
3: \(=\dfrac{3}{4}\cdot\dfrac{7}{5}+\dfrac{9}{7}\cdot\dfrac{3}{2}\)
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
10 tháng 7 2023
1: =-2/9(15/17+2/17)=-2/9
2: =−63+−2190=−63+−2190
=-2-7/30=-67/30
3: =34⋅75+97⋅32=34⋅75+97⋅32
=21/20+27/14=417/140
4: =-25/13(5/19+14/19)=-25/13
5: =-7/5-45/21=-7/5-15/7=-124/35
HN
6
2 tháng 7 2021
1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)
\(x^2-4x+4-x^2+3x-x+3=13\)
\(-2x+7=13\)
\(-2x=13-7\)
\(-2x=6\)
\(x=-3\)
Ta có
\(\left(x-2\right)^2-\left(x+1\right)\left(x-3\right)=13\)
\(\left(x^2-4x+4\right)-\left(x^2-2x-3\right)=13\)
\(-2x+7=13\)
\(x=-3\)
Vậy x=3