`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

\(\Rightarrow2-4x=6-3x\\ \Rightarrow x=-4\)

\(\dfrac{2x-1}{3}=\dfrac{2-x}{-2}\)

\(\Rightarrow-2\left(2x-1\right)=3\left(2-x\right)\)

\(\Rightarrow-4x+2=6-3x\Rightarrow x=-4\)

Có đúng hông dậy mọi người

ĐKXĐ: \(x\ne\left\{-3;-2;-1;0\right\}\)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow\dfrac{3}{x\left(x+3\right)}=\dfrac{x}{x\left(x+3\right)}\)

\(\Leftrightarrow x=3\)

\(a,\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\\ b,\Rightarrow\left(-\dfrac{1}{8}\right)^x=\dfrac{1}{64}=\left(-\dfrac{1}{8}\right)^2\Rightarrow x=2\\ c,\Rightarrow\left(x-2\right)\left(2x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\\ d,\Rightarrow\left(x+1\right)^{x+10}-\left(x+1\right)^{x+4}=0\\ \Rightarrow\left(x+1\right)^{x+4}\left[\left(x+1\right)^6-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}x+1=0\\\left(x+1\right)^6=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x+1=1\\x+1=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=0\\x=-2\end{matrix}\right.\\ e,\Rightarrow\dfrac{3}{4}\sqrt{x}=\dfrac{5}{6}\left(x\ge0\right)\\ \Rightarrow\sqrt{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{100}{81}\)

1.

Điều kiện xác định của căn thức: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{1-1}{1}=0\Rightarrow y=0\) là 1 TCN

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{-1-1}{-1}=2\Rightarrow y=2\) là 1 TCN

\(\lim\limits_{x\rightarrow-5}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{\sqrt{26}+5}{0}=+\infty\Rightarrow x=-5\) là 1 TCĐ

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{\sqrt{26}-5}{0}=+\infty\Rightarrow x=5\) là 1 TCĐ

Hàm có 4 tiệm cận

2.

Căn thức của hàm luôn xác định

Ta có:

\(\lim\limits_{x\rightarrow2}\dfrac{2x-1-\sqrt{x^2+x+3}}{x^2-5x+6}=\lim\limits_{x\rightarrow2}\dfrac{\left(2x-1\right)^2-\left(x^2+x+3\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(3x+1\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{3x+1}{\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}=\dfrac{-7}{6}\) hữu hạn

\(\Rightarrow x=2\) ko phải TCĐ

\(\lim\limits_{x\rightarrow3}\dfrac{2x-1-\sqrt{x^2+x+3}}{x^2-5x+6}=\dfrac{5-\sqrt{15}}{0}=+\infty\)

\(\Rightarrow x=3\) là tiệm cận đứng duy nhất

a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)

b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)

\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)

c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)

\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)

\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)

\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)

\(=\left(\dfrac{3}{7}\right)^{15}\)

\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)

\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)

\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)

\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)

a) 56:55+(49)0=56−5+1=5+1=656:55+(94​)0=56−5+1=5+1=6

b) (37)21:(1−4049)3(73​)21:(1−4940​)3

=(37)21:(949)3=(73​)21:(499​)3

=(37)21:[(37)2]3=(73​)21:[(73​)2]3

=(37)21:(37)6=(73​)21:(73​)6

=(37)21−6=(37)15=(73​)21−6=(73​)15

c) (23)3−(−523)0+49(32​)3−(3−52​)0+94​

=827−1+49=278​−1+94​

=8−27+1227=−727=278−27+12​=−277​
 

1:

a: =>28x-8=9x+3

=>19x=11

=>x=11/19

b: =>(3x-1)(x-1)=(2x+1)(x+1)

=>3x^2-4x+1=2x^2+3x+1

=>x^2-7x=0

=>x=0 hoặc x=7

(5x + 1)² = 36/49

=> (5x + 1)² = (6/7)²

=> \(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{cases}}\)

Làm từ phần b nha

b) \(\left(x-\frac{1}{9}\right)^3=\frac{2}{3}^6\)

\(\Rightarrow\left(x-\frac{2}{9}\right)^3=\left(\frac{1}{3}\right)^6\)

\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1^6}{3^6}\)

\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{3^6}\)

\(\Rightarrow\left(x-\frac{2}{3}\right)^3=\frac{1}{729}\)

\(\Rightarrow x-\frac{2}{9}=\frac{1}{9}\)

      \(x=\frac{1}{9}+\frac{2}{9}\)

      \(x=\frac{3}{9}=\frac{1}{3}\)

c) Sai đề rồi, xem lại đi

d) \(\left(x-3,5\right)^2+\left(y-\frac{1}{10}\right)^4< 0\)

\(\Rightarrow\frac{10000y^4-4000y^3+600y^3-40y+10000x^2+122501-70000x}{10000}< 0\)

=> Sai \(\forall y\inℝ\)

-29 - 9. ( 2x - 1)² = -110