13 - |x + 5| =13
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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
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
\(x=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}}\)
\(\Leftrightarrow x=\sqrt{5+\sqrt{13+x}}\) (\(x\ge0\))
\(\Leftrightarrow x^2=5+\sqrt{13+x}\)
\(\Leftrightarrow x^2-9=\sqrt{13+x}-4\)
\(\Leftrightarrow\left(x-3\right).\left(x+3\right)=\dfrac{x-3}{\sqrt{13+x}+4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=\dfrac{1}{\sqrt{x+13}+4}\left(∗\right)\end{matrix}\right.\)
Xét (*) ta có VT \(\ge3\) (1)
mà \(VP=\dfrac{1}{\sqrt{x+13}+4}\le\dfrac{1}{4}\) (2)
Từ (1) và (2) dễ thấy (*) vô nghiệm
Hay x = 3
a) \(\dfrac{-5}{9}-\dfrac{-5}{12}=\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-20}{36}+\dfrac{15}{36}=-\dfrac{5}{36}\)
b) \(\dfrac{-5}{12}:\dfrac{15}{4}=\dfrac{-5}{12}\times\dfrac{4}{15}=\dfrac{-1}{9}\)
c) \(\dfrac{1}{13}\cdot\dfrac{8}{13}+\dfrac{5}{13}\cdot\dfrac{1}{13}-\dfrac{14}{13}=\dfrac{1}{13}\cdot\left(\dfrac{8}{13}+\dfrac{5}{13}\right)-\dfrac{14}{13}=\dfrac{1}{13}\cdot1-\dfrac{14}{13}=\dfrac{1}{13}-\dfrac{14}{13}=-1\)
`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`
=>125(3x+2):13=1000:13
=>125(3x+2)=1000
=>3x+2=8
=>x=2
\(A.\dfrac{-5}{13}.\dfrac{-4}{13}.\dfrac{-3}{13}.....\dfrac{4}{13}.\dfrac{5}{13}\)
\(A=\dfrac{-5.-4.-3.-2.-1.0.1.2.3.4}{13^{10}}\)
\(A=\dfrac{0}{13^{10}}=0\)
\(A=\dfrac{-5}{13}\cdot\dfrac{-4}{13}\cdot\dfrac{-3}{13}\cdot...\cdot\dfrac{4}{13}\cdot\dfrac{5}{13}\)
\(A=\dfrac{\left(-5\cdot5\right)\cdot\left(-4\cdot4\right)\cdot...\cdot\left(-1\cdot1\right)\cdot0}{13\cdot13\cdot13\cdot...\cdot13}\)
\(A=\dfrac{0}{13\cdot13\cdot13\cdot...\cdot13}\)
\(A=0\)
\(=\dfrac{3\times5\times8\times13}{5\times8\times13\times6}=\dfrac{3}{6}=\dfrac{1}{2}\)
a.11/7.8/13+8/13.3/7+8/13
=8/13.(11/7+3/7+1)
=8/13.3
=24/13
b.5/7.8/13+8/13.9/7-8/13
=8/13.(5/7+9/7-1)
=8/13.1
=8/13
c.157.199-56.199-199
=199.(157-56-1)
=199.100
19900
B=\(\frac{5}{9}\)x{\(\frac{7}{13}\)+\(\frac{9}{13}\)-\(\frac{3}{13}\)}
B=\(\frac{5}{9}\)x 1
B=\(\frac{5}{9}\)
Vậy B=\(\frac{5}{9}\)
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