a) Hàm số đồng biến nếu \(\dfrac{k^2+2}{k-3}>0\) \(\Leftrightarrow k>3\)

b) Hàm số nghịch biến nếu \(\dfrac{k+\sqrt{2}}{k^2+\sqrt{3}}< 0\Leftrightarrow k< -\sqrt{2}\)

\(K=\dfrac{9-5}{3}+\dfrac{2.9-5}{3^2}+\dfrac{3.9-5}{3^3}+...+\dfrac{101.9-5}{3^{101}}\)

\(K=\dfrac{9}{3}+\dfrac{2.9}{3^2}+\dfrac{3.9}{3^3}+...+\dfrac{101.9}{3^{101}}-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)

\(K=9\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\right)-5\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)

\(K=9A-5B\)

Xét \(A=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{101}{3^{101}}\) (1)

\(\Rightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{2}{3^3}+\dfrac{3}{3^4}+...+\dfrac{100}{3^{101}}+\dfrac{101}{3^{102}}\) (2)

Trừ vế với vế (1) cho (2):

\(\dfrac{2}{3}A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}-\dfrac{101}{3^{102}}=B-\dfrac{101}{3^{102}}\)

\(\Rightarrow A=\dfrac{3}{2}\left(B-\dfrac{101}{3^{102}}\right)\Rightarrow K=\dfrac{27}{2}\left(B-\dfrac{101}{3^{102}}\right)-5B\)

\(\Rightarrow K=\dfrac{17}{2}B-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)

Xét \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)

\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{90}}+\dfrac{1}{3^{100}}\)

\(\Rightarrow3B-1+\dfrac{1}{3^{101}}=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}=B\)

\(\Rightarrow2B=1-\dfrac{1}{3^{101}}\Rightarrow B=\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\)

\(\Rightarrow K=\dfrac{17}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}.\dfrac{1}{3^{101}}\right)-\dfrac{27}{2}.\dfrac{101}{3^{102}}\)

\(\Rightarrow K=\dfrac{17}{4}-\dfrac{1}{3^{101}}\left(\dfrac{17}{4}+\dfrac{27.101}{6}\right)< \dfrac{17}{4}\) (đpcm)

Bài 1 :

Để \(\dfrac{x^3+x^2-x-1}{x^3+2x-3}=0\) thì \(x^3+x^2-x-1=0\)

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

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

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)

Vậy,.........

a/c=b/d=k

=>a=ck; b=dk

=>\(\dfrac{c\cdot a^2+d\cdot b^2}{c^3+d^3}\)

\(=\dfrac{c\cdot c^2k^2+d\cdot d^2k^2}{c^3+d^3}=k^2\)

đặt \(\dfrac{a}{c}\) =\(\dfrac{b}{d}=k\)

\(\Rightarrow a=c\times k\)

\(b=d\times k\)

\(\dfrac{c.\left(c.k\right)^2+d.\left(d.k\right)^2}{c^3+d^3}\)

=\(\dfrac{c^3.k^2+d^3.k^2}{c^3+d^3}\)

=\(\dfrac{k^2\left(c^3+d^3\right)}{1\left(c^3+d^3\right)}\)=k²

4 câu đầu hìn như sai đề :v

`m)(3/2-2/(-5)):x-1/2=3/2`

`<=>(3/2+2/5):x=3/2+1/2=2`

`<=>19/10:x=2`

`<=>x=19/10:2=19/20`

`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`

`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`

`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`

`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`

Mà `3/2-5/11-3/13>0`

`<=>2x-2+1/2=0`

`<=>2x-3/2=0`

`<=>2x=3/2<=>x=3/4`

h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)

\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)

\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)

Vậy ...

i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)

\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)

Vậy ...

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12x^2-24-2x=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)

m) Ta có: \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{19}{10}:x=2\)

hay \(x=\dfrac{19}{20}\)

Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10

\(j,\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{53.55}=\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{53}-\dfrac{1}{55}=\dfrac{1}{5}-\dfrac{1}{55}=\dfrac{11}{55}-\dfrac{1}{55}=\dfrac{10}{55}=\dfrac{2}{11}\\ k,\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}...\dfrac{99}{100}=\dfrac{1}{100}.\dfrac{2}{2}.\dfrac{3}{3}...\dfrac{99}{99}=\dfrac{1}{100}.1.1...1=\dfrac{1}{100}\)

d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)

h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)

\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)

k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)

\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)

\(=\dfrac{5}{2}\)

n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)

\(=3-1+\dfrac{1}{8}\cdot16\)

=2+2

=4

1. Sửa đề:

\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)

\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)

2.

\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)

\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)

3.

\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{3(\sqrt{6}+\sqrt{3})}{3}+\frac{4(\sqrt{7}-\sqrt{3})}{4}=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)