26/81 + 4/27= ?
mn giú em nhanh ạ
em cảm ơn!
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1 is going
2 enjoy
3 seem
4 feel
5 are learning
6 needs
7 concentrate
8 support
9 is doing
10 are practicing
11 finds
12 is getting
13 think
1. goes
2. enjoy
3. seem
4. feel
5. are learning
6. needs
7. concentrate
8. support
9. does
10. are practicing
11. finds
12. gets
13. think
\(A=\left(x-3\right)^3-x\left(x^2+27\right)+\left(3x\right)^2\)
\(=x^3-9x^2+27x-27-x^3-27x+9x^2\)
\(=-27\)
\(\Rightarrow\) Giá trị biểu thức A không phụ thuộc biến x.
\(A=\left(x-3\right)^3-x\left(x^2+27\right)+9x^2\)
\(=x^3-9x^2+27x-27-x^3-27x+9x^2\)
=-27
a) \(A=\frac{27\text{x}45+27\text{x}25}{2+4+6+8+...+18}=\frac{27\left(45+25\right)}{\left(18+2\right)\text{x}\left(\frac{18-2}{2}+1\right)}\)
\(A=\frac{27\text{x}70}{20\text{x}9}=\frac{3\text{x}9\text{x}10\text{x}7}{10\text{x}2\text{x}9}\)
\(A=\frac{21}{2}\)
b) \(B=\frac{28\text{x}108-52\text{x}12}{32-28+24-20+16-12+8-4}\)
\(B=\frac{28\text{x}12\text{x}9\text{}\text{}-52\text{x}12}{\left(32-28\right)+\left(24-20\right)+\left(16-12\right)+\left(8-4\right)}\)
\(B=\frac{12\left(28\text{x}9-52\right)}{4+4+4+4}=\frac{12\text{x}200}{4\text{x}4}\)
\(B=\frac{4\text{x}3\text{x}4\text{x}50}{4\text{x}4}=150\)
\(25+27-48-25-37=\left(25-25\right)+\left(27-37\right)-48\)
\(=0-10-48\)
\(=-58\)
\(l,\\ 2^x=1=2^0\\ Vậy:x=0\\ m,\\ 3^x=81=3^4\\ Vậy:x=4\\ n,\\ 3^x=37=3^3\\ Vậy:x=3\\ o,\\ 9^x=3^4=\left(3^2\right)^2=9^2\\ Vậy:x=2\)
a, xét \(\Delta ABC\left(\widehat{BAC}=90^o\right)\) có \(AM\) là đường cao
\(BC^2=AB^2+AC^2\left(pytago\right)\Leftrightarrow BC=\sqrt{12^2+16^2}=20\left(cm\right)\)
\(sinABC=\dfrac{AC}{BC}=\dfrac{16}{20}\Rightarrow\widehat{ABC}\approx53^o8'\)
\(sinACB=\dfrac{AB}{BC}=\dfrac{12}{20}\Rightarrow\widehat{ACB}\approx32^o52'\)
\(AB^2=BM.BC\Rightarrow BM=\dfrac{AB^2}{BC}=\dfrac{12^2}{20}=7,2\left(cm\right)\)
b, Xét \(\Delta ABM\left(\widehat{AMB}=90^o\right)\) có \(AE\perp AB\)
\(AB^2=BM^2+AM^2\left(pytago\right)\Leftrightarrow AM=\sqrt{20^2-7,2^2}=\dfrac{16\sqrt{34}}{5}\left(cm\right)\)
\(AM^2=AE.AB\) (hệ thức lượng trong tam giác vuông)\(\left(1\right)\)
c, Xét \(\Delta AMC\left(\widehat{AMC}=90^o\right)\)
\(AC^2=AM^2+MC^2\left(pytago\right)\Leftrightarrow AM^2=AC^2-MC^2\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow AE.AB=AC^2-MC^2\left(đpcm\right)\)
38/81 nhé
= 26/81 + \(\dfrac{4X3}{27X3}\)
= 26/81 + 12/ 81
= \(\dfrac{26+12}{81}=\dfrac{38}{81}\)