Câu 1:Ta có:

a) \(\left|x-3\right|=5\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

b) \(\left|2x+3\right|=2.\left|4-x\right|\)

+)Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\dfrac{-3}{2}\le x\le4\)

Khi đó \(2x+3=2\left(4-x\right)\Leftrightarrow2x+3=8-2x\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\left(tm\right)\)

+) Xét \(\left\{{}\begin{matrix}2x+3\ge0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow x\ge4\)

Khi đó: \(2x+3=2\left(x-4\right)=2x-8\Leftrightarrow0x=-11\left(vl\right)\)

+) Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\ge0\end{matrix}\right.\) \(\Leftrightarrow x\le\dfrac{-3}{2}\)

Khi đó: \(-\left(2x+3\right)=2.\left(4-x\right)\Leftrightarrow-2x-3=8-2x\left(vl\right)\)

+)Xét \(\left\{{}\begin{matrix}2x+3\le0\\4-x\le0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\le\dfrac{-3}{2}\\x\ge4\end{matrix}\right.\) \(\left(vl\right)\)

Vậy...

c) ĐKXĐ : \(3-x\ge0\Leftrightarrow x\le3\)

+)Xét \(x^{^2}-3x+1\ge0\)

\(\Leftrightarrow x^2-3x+1=3-x\Leftrightarrow x^2-2x-2=0\)

\(\Leftrightarrow x^2-2x+1=3\Leftrightarrow\left(x-1\right)^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\left(tm\right)\\x=1-\sqrt{3}\left(tm\right)\end{matrix}\right.\)

+)Xét \(x^{^2}-3x+1\le0\)

\(\Leftrightarrow-\left(x^2-3x+1\right)=3-x\)

\(\Leftrightarrow x^2-3x+1=x-3\Leftrightarrow x^2-4x+4=0\)

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

Vậy...

Câu 2:

 Ta có:

Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có một nghiệm là \(x=-3\)

\(\Rightarrow\)Phương trình \(\left(x+3\right)\left(x^2-2x+m-1\right)=0\) có ba nghiệm phân biệt khi và chỉ khi \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\)

Ta có:  \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt khi và chỉ khi \(\text{△}>0\Leftrightarrow8-4m>0\Leftrightarrow m< 2\)

 Gọi \(x_1\) và \(x_2\) là 2 nghiệm của phương trình \(x^2-2x+m-1=0\).Theo hệ thức Vi-ét ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{-2}{1}=2\\x_1x_2=\dfrac{m-1}{1}=m-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2-x_2\\\left(2-x_2\right).x_2=m-1\end{matrix}\right.\)

Nếu \(x_2\ne-3\) thì \(m-1\ne-15\Leftrightarrow m\ne-14\).

Do vai trò của  \(x_1\) và \(x_2\) là như nhau nên  \(x^2-2x+m-1=0\) có 2 nghiệm phân biệt và khác \(-3\) khi và chỉ khi: \(\left\{{}\begin{matrix}m< 2\\m\ne-14\end{matrix}\right.\)

Câu 2: 

\(C=-x+\sqrt{x}\)

\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)

1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

rối qué với cả vì hum bt

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=1-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

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

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

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

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

h) \(x-\dfrac{2}{20}=-\dfrac{5}{2}-x\)

\(\Rightarrow x+x=-\dfrac{5}{2}+\dfrac{2}{20}\)

\(\Rightarrow2x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}:2=-\dfrac{6}{5}\)

i) \(\left(\dfrac{x}{2}-1\right)^3+2=-\dfrac{11}{8}\)

\(\Rightarrow\left(\dfrac{x}{2}-1\right)^3=-\dfrac{11}{8}-2\)

\(\Rightarrow\dfrac{x}{2}-1=\sqrt[3]{-\dfrac{27}{8}}\)

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

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

\(\Rightarrow x=-\dfrac{1}{2}.2=-1\)

k) \(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{3}{4}-1\dfrac{1}{2}x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}.3=-\dfrac{3}{2}\\x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\end{matrix}\right.\)

1.

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

2.

a. \(y'=6x^2-sinx-\dfrac{1}{2\sqrt{x}}\)

b. \(y'=10\left(x^2-5\right)^9.\left(x^2-5\right)'=20x\left(x^2-5\right)^9\)

3.

\(y'=-2x\)

\(k=4\Rightarrow-2x=4\Rightarrow x=-2\Rightarrow y\left(-2\right)=-24\)

Phương trình tiếp tuyến:

\(y=4\left(x+2\right)-24\Leftrightarrow y=4x-16\)

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

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

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

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

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

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)