gúp mình nhanh nha

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\(\frac{3^9-2^3.3^7+2^{10}.3^2-2^{13}}{3^{10}-2^2.3^7+2^{10}.3^2-2^{12}}\)\(=\frac{2-2}{3}\)\(=\frac{0}{3}=0\)

1 mũ 3= 3 mũ ko biết

làm gì có 1 mũ 3 = 3

Bài làm:

Ta có: \(\frac{AB}{AC}=\frac{3}{4}\Rightarrow AB=\frac{3}{4}AC\)

Vì tam giác ABC vuông tại A nên theo định lý Pytago ta có:

\(AB^2+AC^2=BC^2\)

\(\Leftrightarrow\frac{9}{16}AC^2+AC^2=100\)

\(\Leftrightarrow\frac{25}{16}AC^2=100\Leftrightarrow AC^2=64\Rightarrow AC=8\left(cm\right)\Rightarrow AB=\frac{3}{4}AC=6\left(cm\right)\)

Lại có: \(AB\cdot AC=AH\cdot BC\left(=2S_{ABC}\right)\)

\(\Leftrightarrow6\cdot8=10AH\Leftrightarrow AH=\frac{6\cdot8}{10}=\frac{24}{5}\left(cm\right)\)

Vậy AH = 24/5(cm)

Xét \(\Delta ABC\) vuông tại A có:

\(BC^2=AB^2+AC^2\) (định lí Pytago)

\(\Rightarrow AB^2+AC^2=10^2=100\)

Ta có: \(AB:AC=3:4\Rightarrow\frac{AB}{3}=\frac{AC}{4}\Rightarrow\frac{AB^2}{9}=\frac{AC^2}{16}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{AB^2}{9}=\frac{AC^2}{16}=\frac{AB^2+AC^2}{9+16}=\frac{100^2}{25}=4\)

\(\Rightarrow\hept{\begin{cases}\frac{AB^2}{9}=4\\\frac{AC^2}{16}=4\end{cases}}\Rightarrow\hept{\begin{cases}AB^2=36\\AC^2=64\end{cases}}\Rightarrow\hept{\begin{cases}AB=6\left(cm\right)\\AC=8\left(cm\right)\end{cases}}\) (vì \(AB,AC>0\))

Ta có: \(S_{\Delta ABC}=\frac{AB.AC}{2}=\frac{AH.BC}{2}\)

\(\Rightarrow AB.AC=AH.BC\)

hay \(6.8=10AH\)

\(\Rightarrow AH=\frac{6.8}{10}=4,8\left(cm\right)\)

Vậy \(AH=4,8cm\).

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\) 

\(=\frac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Rightarrow\hept{\begin{cases}a+b+c=3a\\a+b+c=3b\\a+b+c=3c\end{cases}}\Rightarrow a=b=c\)

Khi đó: \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=2^3=8\)

Vậy B = 8

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(\Leftrightarrow\frac{a+b}{c}-1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)

\(\Leftrightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)

\(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(B=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)

\(B=\frac{a+b}{c}.\frac{c+a}{b}.\frac{b+c}{a}=2.2.2=8\)

*Chứng tỏ \(x=\frac{1}{2}\) là nghiệm của đa thức \(P\left(x\right)=4x^2-4x+1\)

Cho \(P\left(x\right)=0\)

\(\Rightarrow4x^2-4x+1=0\)

\(\Rightarrow4x^2-2x-2x+1=0\)

\(\Rightarrow2x\left(2x-1\right)-\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)\left(2x-1\right)=0\)

\(\Rightarrow\left(2x-1\right)^2=0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=\frac{1}{2}\)

\(\Rightarrow P\left(x\right)\) có nghiệm là \(x=\frac{1}{2}\)

\(\Rightarrowđpcm\)

*Chứng tỏ đa thức \(Q\left(x\right)=4x^2+1\) không có nghiệm

Ta có: \(4x^2\ge0\forall x\)

\(\Rightarrow4x^2+1>0\)

hay \(Q\left(x\right)>0\)

\(\Rightarrow\)Đa thức \(Q\left(x\right)=4x^2+1\) không có nghiệm   (đpcm)

chịu :))))

.... :)))))))))

\(A=\frac{1}{5}+\frac{1}{5^2}+...............+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\)

\(\Rightarrow5A=1+\frac{1}{5}+...................+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\)

\(\Rightarrow5A-A=\left(1+\frac{1}{5}+...........+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+.............+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)\)

\(\Rightarrow4A=1-\frac{1}{5^{2017}}\)

\(\Rightarrow A=\left(1-\frac{1}{5^{2017}}\right):4=\left(1-\frac{1}{5^{2017}}\right).\frac{1}{4}=\frac{1}{4}-\frac{1}{5^{2017}.4}< \frac{1}{4}\)

\(\Rightarrow A< \frac{1}{4}\)

Vậy \(A< \frac{1}{4}\)

Chúc bạn học tốt

\(A=\frac{1}{5}+\frac{1}{5^2}+....+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\)

=>5A = \(1+\frac{1}{5}+...+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\)

=> 5A -A = \(\left(1+\frac{1}{5}+...+\frac{1}{5^{2015}}+\frac{1}{5^{2016}}\right)-\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{2016}}+\frac{1}{5^{2017}}\right)\)

=> 4A = \(1-\frac{1}{5^{2017}}\)

=> \(A=\frac{1-\frac{1}{5^{2017}}}{4}=\frac{1}{4}-\frac{1}{5^{2017}}< \frac{1}{4}\)