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6:
\(2^{225}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=\left(3^2\right)^{75}=9^{75}\)
mà 8<9
nên \(2^{225}< 3^{150}\)
4: \(\left|5x+3\right|>=0\forall x\)
=>\(-\left|5x+3\right|< =0\forall x\)
=>\(-\left|5x+3\right|+5< =5\forall x\)
Dấu = xảy ra khi 5x+3=0
=>x=-3/5
1:
\(\left(2x+1\right)^4>=0\)
=>\(\left(2x+1\right)^4+2>=2\)
=>\(M=\dfrac{3}{\left(2x+1\right)^4+2}< =\dfrac{3}{2}\)
Dấu = xảy ra khi 2x+1=0
=>x=-1/2
\(c,-\dfrac{8}{13}+\left(-\dfrac{7}{5}-x\right)=-\dfrac{1}{2}\\ -\dfrac{7}{5}-x=-\dfrac{1}{2}-\dfrac{8}{13}\\ -\dfrac{7}{5}-x=-\dfrac{29}{26}\\ x=-\dfrac{7}{5}-\left(-\dfrac{29}{26}\right)=-\dfrac{37}{130}\\ d,-1\dfrac{1}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{8}{7}-\left[-\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)\right]=-\dfrac{4}{21}\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{8}{7}-\left(-\dfrac{4}{21}\right)\\ -\dfrac{5}{3}+\left(x-\dfrac{7}{3}\right)=-\dfrac{20}{21}\\ x-\dfrac{7}{3}=-\dfrac{20}{21}-\left(-\dfrac{5}{3}\right)\\ x-\dfrac{7}{3}=\dfrac{5}{7}\\ x=\dfrac{5}{7}+\dfrac{7}{3}=\dfrac{64}{21}\\ e,-\dfrac{2}{3}-x:\dfrac{1}{2}=\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{2}{3}-\dfrac{2}{5}\\ x:\dfrac{1}{2}=-\dfrac{16}{15}\\ x=-\dfrac{16}{15}\times\dfrac{1}{2}=-\dfrac{8}{15}\)
c: -8/13+(-7/5-x)=-1/2
=>x+7/5+8/13=1/2
=>x=1/2-7/5-8/13=-197/130
d: \(\Leftrightarrow-\dfrac{8}{7}+\dfrac{5}{3}-\left(x-\dfrac{7}{3}\right)=\dfrac{-4}{21}\)
=>\(x-\dfrac{7}{3}=\dfrac{-8}{7}+\dfrac{5}{3}+\dfrac{4}{21}=\dfrac{-24+35+4}{21}=\dfrac{18}{21}=\dfrac{6}{7}\)
=>x=6/7+7/3=18/21+49/21=67/21
e: =>x:1/2=-2/3-2/5=-16/15
=>x=-16/15*1/2=-8/15
f: =>-8/5*x=-1/3+4/9=1/9
=>x=-1/9:8/5=-1/9*5/8=-5/72
g: =>-4/5x-1/4+x=-13/3
=>1/5x=-13/3+1/4=-52/12+3/12=-49/12
=>x=-49/12*5=-245/12
h: =>12/7:x-1/2=0 hoặc 2/5x-3/2=0
=>12/7:x=1/2 hoặc 2/5x=3/2
=>x=12/7:1/2=24/7 hoặc x=3/2:2/5=3/2*5/2=15/4
a) Ta có:
\(\widehat{yOu}+\widehat{xOy}=180^o\) (kề bù)
\(\Rightarrow\widehat{yOu}=180^o-\widehat{xOy}\)
\(\Rightarrow\widehat{yOu}=180^o-60^o=120^o\)
Mà: \(\widehat{xOt}+\widehat{tOu}=180^o\) (kề bù)
\(\Rightarrow\widehat{xOt}=180^o-\widehat{tOu}\)
\(\Rightarrow\widehat{xOt}=180^o-30^o=150^o\)
b) Ta có:
\(\widehat{xOy}+\widehat{yOt}+\widehat{tOu}=\widehat{xOu}\)
\(\Rightarrow\widehat{yOt}=\widehat{xOu}-\widehat{xOy}-\widehat{tOu}\)
\(\Rightarrow\widehat{yOt}=180^o-60^o-30^o\)
\(\Rightarrow\widehat{yOt}=90^o\)
1: \(\sqrt{11}\) là số vô tỉ
2:
a: 4,9(18)=4,91818...
mà 4,91818<4,928
nên 4,9(18)<4,928
b: 4,315<4,318
=>-4,315>-4,318
=>-4,315...>-4,318...
c: \(\sqrt{3}=\sqrt{\dfrac{6}{2}}< \sqrt{\dfrac{7}{2}}\)
3:
a: \(6=\sqrt{3};-1,7=-\sqrt{2,89}\)
0<2,89<3
=>\(0< \sqrt{2,89}< \sqrt{3}\)
=>\(-\sqrt{3}< -\sqrt{2,89}< 0\)
0<35<36<47
=>\(0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)
=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< \sqrt{36}< \sqrt{47}\)
=>\(-\sqrt{3}< -\sqrt{2,89}< 0< \sqrt{35}< 6< \sqrt{47}\)
b: \(-\sqrt{2\dfrac{1}{3}}=-\sqrt{2,\left(3\right)}\)
\(-1,5=-\sqrt{2,25}\)
2,25<2,3<2,(3)
=>\(\sqrt{2.25}< \sqrt{2.3}< \sqrt{2.\left(3\right)}\)
=>\(0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)
\(0< \sqrt{5\dfrac{1}{6}}=\sqrt{5,1\left(6\right)}< \sqrt{5,3}\)
=>\(\sqrt{5.3}>\sqrt{5\dfrac{1}{6}}>0>-1.5>-\sqrt{2.3}>-\sqrt{2\dfrac{1}{3}}\)
`#1231.2021`
`1.`
Ta có:
`y` tỉ lệ nghịch với `x` theo hệ số tỉ lệ `-4`
`=> y = (-4)/x` `(1)`
`x` tỉ lệ nghịch với `z` theo hệ số tỉ lệ `3/4`
`=> x = 3/4 \div z` `(2)`
Thay `(2)` vào `(1)`
`=> y = (-4)/(3/4 \div z) => y = -16/3 * z`
Vậy, `y` và `z` tỉ lệ thuận với nhau theo hệ số tỉ lệ `-16/3`
`=> A.`
\(a,x+\dfrac{1}{2}=\dfrac{3}{4}\\ x=\dfrac{3}{4}-\dfrac{1}{2}\\ x=\dfrac{1}{2}\\ b,-\dfrac{2}{3}-x=1\\x=-\dfrac{2}{3}-1\\ x=-\dfrac{5}{3}\\ d,\dfrac{1}{4}+\dfrac{3}{4}:x=\dfrac{5}{2}\\ \dfrac{3}{4}:x=\dfrac{5}{2}-\dfrac{1}{4}\\ \dfrac{3}{4}:x=\dfrac{9}{4}\\ x=\dfrac{3}{4}:\dfrac{9}{4}\\ x=\dfrac{1}{3}\\ e,\left(x+\dfrac{1}{4}\right)\cdot\dfrac{3}{4}=-\dfrac{5}{8}\\ x+\dfrac{1}{4}=-\dfrac{5}{8}:\dfrac{3}{4}\\ x+\dfrac{1}{4}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{1}{4}\\ x=\dfrac{7}{12}\)
\(g,\dfrac{x-3}{15}=\dfrac{-2}{5}\\ 5\left(x-3\right)=-30\\ x-3=-6\\ x=-6+3\\ x=-3\\ h,\dfrac{x}{-2}=\dfrac{-8}{x}\\ x^2=16\\ x=\pm\sqrt{16}\\ x=\pm4\\ k,\dfrac{x+2}{3}=\dfrac{x-4}{5}\\ 5\left(x+2\right)=3\left(x-4\right)\\ 5x+10=3x-12\\ 5x-3x=-12-10\\ 2x=-22\\ x=-11\)
\(m,\left(2x-1\right)^2=4\\ \Rightarrow\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
2:
a: Xét ΔABD và ΔACD có
AB=AC
BD=CD
AD chung
=>ΔABD=ΔACD
b: ΔABD=ΔACD
=>góc ADB=góc ADC=180/2=90 độ
=>AD vuông góc BC
c: BC=12
=>BD=CD=6
AD=căn 10^2-6^2=8
d: BN=AB/2
CM=AC/2
mà AB=AC
nên BN=CM
Xét ΔNBC và ΔMCB có
NB=MC
góc NBC=góc MCB
BC chung
=>ΔNBC=ΔMCB
=>NC=BM
e: Xét ΔABC có
BM,CN là trung tuyến
BM cắt CN tại G
=>G là trọng tâm
=>A,G,D thẳng hàng và AG=2/3AD=16/3
Bài 16
a) \(A=\dfrac{n+1}{n+2}\)
Gọi ƯCLN(n+1;n+2) là x ( \(x\in N\) *)
\(\Rightarrow\) \(\left\{{}\begin{matrix}\left(n+1\right)⋮x\\\left(n+2\right)⋮x\end{matrix}\right.\)
\(\Rightarrow\) \(\left(n+2\right)-\left(n+1\right)\) \(⋮x\)
\(\Rightarrow\) \(1\) \(⋮x\)
\(\Rightarrow\) x = 1 \(\Rightarrow\) ƯCLN(n+1;n+2)=1
Vậy A là phân số tối giản ( vì có ƯCLN = 1)
b) \(B=\dfrac{n+1}{3n+4}\)
Gọi ƯCLN(n+1;3n+4) là d ( \(d\in N\) *)
\(\Rightarrow\) \(\left\{{}\begin{matrix}n+1⋮d\\3n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}3n+3⋮d\\3n+4⋮d\end{matrix}\right.\)
\(\Rightarrow\) (3n+4)-(3n+3) chia hết cho d
\(\Rightarrow\) \(1⋮d\)
\(\Rightarrow\) d =1
Vậy B là phân số tối giản.
Mấy phần kia tương tự
c: Gọi d=ƯCLN(3n+2;5n+3)
=>3n+2 chia hết cho d và 5n+3 chia hết cho d
=>15n+10 chia hết cho d và 15n+9 chia hết cho d
=>1 chia hết cho d
=>ƯCLN(3n+2;5n+3)=1
=>PSTG
d: Gọi d=ƯCLN(12n+1;30n+2)
=>12n+1 và 30n+2 đều chia hết cho d
=>60n+5 chia hết cho d và 60n+4 chia hết cho d
=>1 chia hết cho d
=>d=1
=>PSTG
38 : 81 = 38 : 34 = 38-4 = 34.
\(\frac{3^8}{81}\)
\(=\frac{3^8}{3^4}\)
\(=3^{8-4}\)
\(=3^4\)