1.

\(y'=12x+\dfrac{4}{x^2}\)

2.

\(y'=\dfrac{3}{\left(-x+1\right)^2}\)

3.

\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)

4.

\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)

\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)

5.

\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)

6.

\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)

\(\text{a) }\left(x-1\right)\left(x-5\right)>0\\ \text{ Để }\left(x-1\right)\left(x-5\right)>0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ cùng dấu }\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên dương }\Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5>0\Rightarrow x>5\end{matrix}\right.\Rightarrow x>5\\ \text{+) Xét }x-1\text{ và }x-5\text{ là số nguyên âm }\Rightarrow\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow x< 1\\ \text{Vậy }\left(x-1\right)\left(x-5\right)>0\text{ khi }x< 1\text{ hoặc }x>5\)

\(\text{b) }\left(x-1\right)\left(x-5\right)< 0\\ \text{ Để }\left(x-1\right)\left(x-5\right)< 0\text{ thì }\Rightarrow x-1\text{ và }x-5\text{ trái dấu }\\ \text{ Mà }x-1>x-5\\ \Rightarrow\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x-5< 0\Rightarrow x< 5\end{matrix}\right.\Rightarrow1< x< 5\\ \text{ Vậy }\left(x-1\right)\left(x-5\right)< 0\text{ khi }1< x< 5\)

\(\text{c) }\dfrac{3}{4}-\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{4}\\ \Leftrightarrow\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{2}\\ \Leftrightarrow\left|x-\dfrac{1}{7}\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{7}=-2\\x-\dfrac{1}{7}=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{7}\\x=\dfrac{15}{7}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{13}{7}\text{ hoặc }x=\dfrac{15}{7}\)

\(\text{d) }\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=-\dfrac{1}{4}\\x-\dfrac{1}{2}=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{1}{4}\text{ hoặc }x=\dfrac{3}{4}\)

\(\text{e) }8\left(x+1\right)-2\left(2x+5\right)=0\\ \Leftrightarrow8x+8-4x+10=0\\ \Leftrightarrow\left(8x-4x\right)+\left(8+10\right)=0\\ \Leftrightarrow4x+18=0\\ \Leftrightarrow4x=-18\\ \Leftrightarrow x=-\dfrac{9}{2}\\ \text{Vậy }x=-\dfrac{9}{2}\)

\(\text{g) }\left(6x-1\right)-\left(x+8\right)=0\\ \Leftrightarrow6x-1-x-8=0\\ \Leftrightarrow\left(6x-x\right)-\left(1+8\right)=0\\ \Leftrightarrow5x-9=0\\ \Leftrightarrow5x=9\\ \Leftrightarrow x=\dfrac{9}{5}\\ \text{Vậy }x=\dfrac{9}{5}\)

\(\text{h) }\left|7x-\dfrac{1}{4}\right|=1\\ \Leftrightarrow\left[{}\begin{matrix}7x-\dfrac{1}{4}=-1\\7x-\dfrac{1}{4}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=-\dfrac{3}{4}\\7x=\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{28}\\x=\dfrac{5}{28}\end{matrix}\right.\\ \text{Vậy }x=-\dfrac{3}{28}\text{ hoặc }x=\dfrac{5}{28}\)

\(\text{q) }-2x-3=-x+7\\ \Leftrightarrow-2x-3-\left(-x+7\right)=0\\ \Leftrightarrow-2x-3+x-7=0\\ \Leftrightarrow\left(-2x+x\right)-\left(3+7\right)=0\\ \Leftrightarrow-x-10=0\\ \Leftrightarrow-x=10\\ \Leftrightarrow x=-10\\ \text{ Vậy }x=-10\)

mình chưa học tới nhé

a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)

\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)

c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)

\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)

d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)

\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)

e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)

\(minE=-20\Leftrightarrow x=-3\)

f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)

\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)

a: Ta có: \(A=x^2+3x+4\)

\(=x^2+2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}+\dfrac{7}{4}\)

\(=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{3}{2}\)

d: Ta có: \(D=4x^2+4x-24\)

\(=4x^2+4x+1-25\)

\(=\left(2x+1\right)^2-25\ge-25\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{1}{2}\)

e: ta có: \(E=x^2+6x-11\)

\(=x^2+6x+9-20\)

\(=\left(x+3\right)^2-20\ge-20\forall x\)

Dấu '=' xảy ra khi x=-3

vâng ạ

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

<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)

<=>9x-6=2-4x

<=>9x+4x=2+6

<=>13x=8

<=>x=\(\dfrac{8}{13}\)

1.a)2(x-0,5)+3=0,25(4x-1)

<=>2x-1+3=x-1phần4

<=>2x-x=-1/4+1-3

<=>x=-3/4

a. 2( x - 0,5 ) + 3 = 0,25(4x - 1 )

(=) 2x - 1 + 3 = x - 0,25

(=) 2x - x = 1 - 3 - 0,25

(=) x = -2,25

b. 2.

(=) 2x - 2x = 0,5 + 4 -3 +2 ( vô lí )

Vậy phương trình này vô nghiệm

1,

x¹⁰ = x

=> x¹⁰ - x = 0

=> x(x⁹ - 1) = 0

=> \(\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x^9=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

KL: x thuộc {1; 0}

2,

\(S=2+2^2+2^3+...+2^{2016}\)

=> \(2S=2^2+2^3+2^4+...+2^{2017}\)

=> \(2S-S=\left(2^2+2^3+2^4+...+2^{2017}\right)-\left(2+2^2+2^3+...+2^{2016}\right)\)

=> \(S=2^{2017}-2\)

Bài 1:

x¹⁰ = x => x= { -1;1}

Bài 2:

\(S=2+2^2+2^3+...+2^{2016}\)

\(2S=2^2+2^3+2^4+2^{2017}\)

\(2S-S=2^{2017}-2\)

Vậy \(S=2^{2017}-2\)

2.Tính: S= 2 + 2² + 2³ + ....... + 2²⁰¹⁶

=> 2S = 2² + 2³ + ....... + 2²⁰¹⁶ 

=> 2S - S = 2²⁰¹⁶ - 2

=> S = 2²⁰¹⁶ - 2

a/\(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)
\(x=1+\dfrac{5}{7}+\dfrac{13}{14}\)
\(x=\dfrac{14}{14}+\dfrac{10}{14}+\dfrac{13}{14}\)
\(x=\dfrac{37}{14}\)
Vậy \(x=\dfrac{37}{14}\)
b/\(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)
\(x+\dfrac{3}{5}+\dfrac{6}{5}=\dfrac{11}{3}\)
\(x+\dfrac{9}{5}=\dfrac{11}{3}\)
\(x=\dfrac{11}{3}-\dfrac{9}{5}\)
\(x=\dfrac{55}{15}-\dfrac{27}{15}\)
\(x=\dfrac{28}{15}\)
Vậy \(x=\dfrac{28}{15}\)
#kễnh

a) \(x-\dfrac{5}{7}-\dfrac{13}{14}=1\)

\(x-\dfrac{23}{14}=1\)

\(x=1+\dfrac{23}{14}\)

\(x=\dfrac{37}{14}\)

b) \(\dfrac{3}{5}+x+1\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{11}{3}\)

\(x+\dfrac{9}{5}=\dfrac{11}{3}\)

\(x=\dfrac{11}{3}-\dfrac{9}{5}\)

\(x=\dfrac{28}{15}\)

a) ĐKXĐ: \(x\ne\pm2\)

b) \(A=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)\cdot\dfrac{x^2-4x+4}{4}\)

\(=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)^2}{4}\)

\(=\dfrac{4\left(x-2\right)^2}{4\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x-2}{x+2}\)

c) Với \(x=4\) thoả mãn điều kiện \(x\ne\pm2\), nên thay \(x=4\) vào A, ta có:

\(A=\dfrac{4-2}{4+2}=\dfrac{2}{6}=\dfrac{1}{3}\)

a) A xác định \(\Leftrightarrow\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

b) \(A=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right)\cdot\dfrac{x^2-4x+4}{4}\)

\(A=\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)^2}{4}\)

\(A=\dfrac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

\(A=\dfrac{x-2}{x+2}\)

c) Thay x = 4 ( thỏa mãn ĐKXĐ ), ta có :

\(A=\dfrac{4-2}{4+2}=\dfrac{2}{5}=\dfrac{1}{3}\)