Unruly Kid Rr :))

:))

a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)

=>(2x-1)(x-2)(x+1)<>0

hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)

b: ĐKXĐ: x+5<>0

=>x<>-5

c: ĐKXĐ: x⁴-1<>0

hay \(x\notin\left\{1;-1\right\}\)

d: ĐKXĐ: \(x^4+2x^2-3< >0\)

=>\(x\notin\left\{1;-1\right\}\)

$a)\frac{2x}{2x^{2}-5x+3}+\frac{13x}{2x^{2}+x+3}=6$ (1)

Nhận thấy x=0 ko phải nghiệm của phương trình

Chia cả tử và mẫu của mỗi phân thức cho x, ta được:

$\frac{2}{2x-5+\frac{3}{x}}+\frac{13}{2x+1+\frac{3}{x}}=6$

Đặt $2x+\frac{3}{x}$=t

=> (1) <=> $\frac{2}{t-5}+\frac{13}{t+1}=6$

<=> $2t^{2}-13t+11=0$

Có a+b+c=2-13+11=0

=> $t_{1}=1$

$t_{2}=\frac{c}{a}=\frac{11}{2}$

* t = 1

=> $2x+\frac{3}{x}=1$

<=> $2x^{2}-x+3=0$ (vô nghiệm)

* t = $\frac{11}{2}$

=> $2x+\frac{3}{x}=\frac{11}{2}$

<=> $4x^{2}-11x+6=0$

=> $x_{1}=\frac{3}{4}$

$x_{2}=2$

Vậy phương trình có tập nghiệm S={$\frac{3}{4};2$}

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

\(\Leftrightarrow\left[x^2+\left(\dfrac{x}{x-1}\right)^2+2.x.\dfrac{x}{x-1}\right]-2.\dfrac{x^2}{x-1}-1=0\)

\(\Leftrightarrow\left(x+\dfrac{x}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\)

\(\Leftrightarrow\left(\dfrac{x\left(x-1\right)+x}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\)

\(\Leftrightarrow\left(\dfrac{x^2}{x-1}\right)^2-2.\dfrac{x^2}{x-1}-1=0\) (1)

Đặt : \(\dfrac{x^2}{x-1}=t\) (*) thì phương trình (1) trở thành:

\(t^2-2t-1=0\)

Ta có: \(\Delta=8>0\)

\(\Rightarrow t_1=\dfrac{2-\sqrt{8}}{2}=\dfrac{2-2\sqrt{2}}{2}=1-\sqrt{2}\)

\(t_2=\dfrac{2+\sqrt{8}}{2}=\dfrac{2+2\sqrt{2}}{2}=1+\sqrt{2}\)

Thay vào (*) rồi tìm x là xong

=.= hk tốt!!

a) để \(y=\sqrt{x+6\sqrt{x-1}+8}+\dfrac{5}{1-x}\) có nghĩa

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\1-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x\ne1\end{matrix}\right.\Rightarrow x>1\) vậy \(x>1\)

b) để \(y=\dfrac{3x-5}{x^3-x^2+3x-3}\) có nghĩa

\(\Leftrightarrow x^3-x^2+3x-3\ne0\Leftrightarrow x^2\left(x-1\right)+3\left(x-1\right)\ne0\)

\(\Leftrightarrow\left(x^2+3\right)\left(x-1\right)\ne0\Leftrightarrow x-1\ne0\Leftrightarrow x\ne1\)

c) để \(y=\dfrac{3x+1}{\left|3x-1\right|+\left|x-7\right|}\ne0\)

\(\Leftrightarrow\left|3x-1\right|+\left|x-7\right|\ne0\Leftrightarrow\left[{}\begin{matrix}3x-1\ne0\\x-7\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{1}{3}\\x\ne7\end{matrix}\right.\)

\(\Rightarrow x\in R\)

d) để : \(y=\dfrac{\sqrt{x-2}}{\left|x-3\right|+\sqrt{9-x^2}}\) có nghĩa

\(\Leftrightarrow\left\{{}\begin{matrix}x-2\ge0\\9-x^2\ge0\\\left|x-3\right|+\sqrt{9-x^2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\-3\le x\le3\\x\ne3\end{matrix}\right.\Rightarrow2\le x< 3\)

1: ĐKXĐ: \(\left|x^2-4\right|+\left|x+2\right|< >0\)

\(\Leftrightarrow x\ne-2\)

2: ĐKXĐ: \(\left|x-2\right|-\left|x+1\right|< >0\)

\(\Leftrightarrow\left|x-2\right|< >\left|x+1\right|\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2< >x+1\\x-2< >-x-1\end{matrix}\right.\Leftrightarrow2x< >1\Leftrightarrow x< >\dfrac{1}{2}\)

3: ĐKXĐ: \(\left\{{}\begin{matrix}2x+11>=0\\\left\{{}\begin{matrix}3x-2< >4\\3x-2< >-4\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{11}{2}\\x\notin\left\{2;-\dfrac{2}{3}\right\}\end{matrix}\right.\)

Bài 2: 

a: \(A=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}\)

\(=4-\dfrac{4}{7}=\dfrac{24}{7}\)

b: \(B=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}\)

\(=5+\dfrac{7}{11}=\dfrac{62}{11}\)

c: \(C=\dfrac{-5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+1+\dfrac{5}{7}=1\)

d: \(D=\dfrac{7}{10}\cdot\dfrac{8}{3}\cdot20\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}\)

\(=\dfrac{20}{10}\cdot7\cdot\dfrac{8}{3}\cdot\dfrac{3}{8}\cdot\dfrac{5}{28}=2\cdot\dfrac{5}{4}=\dfrac{5}{2}\)

\(f\left(-2\right)-f\left(1\right)=\left(-2\right)^2+2+\sqrt{2-\left(-2\right)}-\left(1^2+2+\sqrt{2-1}\right)\) \(=8-4=4\).
\(f\left(-7\right)-g\left(-7\right)=\left(-7\right)^2+2+\sqrt{2-\left(-7\right)}-\left(-2.\left(-7\right)^3-3.\left(-7\right)+5\right)=-658\)

a)TXĐ:\(x-1>0\Rightarrow x>1\)