Làm mẫu 1 phần :

a) \(|3x-1|+|x-1|=4\left(1\right)\)

Ta có: \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

             \(x-1=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

+) Với \(x< \frac{1}{3}\Rightarrow\hept{\begin{cases}3x-1< 0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=1-3x\\|x-1|=1-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(1-3x\right)+\left(1-x\right)=4\)

\(2-4x=4\)

\(4x=-2\)

\(x=\frac{-1}{2}\)( chọn )

+) Với \(\frac{1}{3}\le x< 1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=1-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :
\(\left(3x-1\right)+\left(1-x\right)=4\)

\(2x=4\)

\(x=2\)( chọn )

+) Với \(x\ge1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1>0\end{cases}\Rightarrow}\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=x-1\end{cases}\left(4\right)}\)

Thay (4) vào (1) ta được :

\(\left(3x-1\right)+\left(x-1\right)=4\)

\(4x-2=4\)

\(4x=6\)

\(x=\frac{3}{2}\)( chọn )

Vậy \(x\in\left\{\frac{-1}{2};2;\frac{3}{2}\right\}\)

a) <=? |(x-1/4)| = 1/4-x

Th1: x >= 1/4 => x - 1/4 = 1/4 - x

<=> 2x = 2.1/4 <=> x = 1/4(nhân)

Th2: x<1/4 => -x + 1/4 = 1/4-x

<=> 0x = 0

<=> x thuộc R và x <1/4.

Vậy S ={x|x<=1/4}

\(\text{a)}\sqrt{x^2-\frac{1}{2}x+\frac{1}{16}}=\frac{1}{4}-x\)

\(\Leftrightarrow\sqrt{x^2-2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2}=\frac{1}{4}-x\)

\(\Leftrightarrow\sqrt{\left(x-\frac{1}{4}\right)^2}=\frac{1}{4}-x\)

\(\Leftrightarrow x-\frac{1}{4}=\frac{1}{4}-x\)

\(\Leftrightarrow2x=\frac{1}{2}\)

\(\Leftrightarrow x=\frac{1}{4}\)

\(\text{b)}\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}-1\)

\(ĐKXĐ:x\ge-2\)

\(\Leftrightarrow\left(\sqrt{x-2\sqrt{x-1}}\right)^2=\left(\sqrt{x-1}-1\right)^2\)

\(\Leftrightarrow x-2\sqrt{x-1}=\left(\sqrt{x-1}\right)^2-2\sqrt{x-1}+1\)

\(\Leftrightarrow x-2\sqrt{x-1}=x-1-2\sqrt{x-1}+1\)

\(\Leftrightarrow x-2\sqrt{x-1}-x+2\sqrt{x-1}=-1+1\)

\(\Leftrightarrow0x=0\)

Vậy \(S=\left\{x\inℝ|x\ge-2\right\}\)

viết lộn nha x^2+2x+1/x^4-2x^2+2

1,44224957+2,080083823=3,522333393 \(\in\)I

Liên quan gì bạn @Tam Mai, chứng minh chứ không phải bấm máy tính

\(p=\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)

\(=\frac{\sqrt{x}.\sqrt{x}-1}{\sqrt{x}}:\left(\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)

\(=\frac{x-1}{\sqrt{x}}:\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}.\frac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}.\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)

Vy Thị Hoàng Lan: sao dòng thứ 3 lại biến đổi đc sang dòng thứ 4 vậy ạ

Mh chưa hiểu mong giải đáp nha

Mơnn bn n~~

\(P=\frac{2n^2-n+2}{2n+1}=\frac{n\left(2n+1\right)-\left(2n-2\right)}{2n+1}=n-\frac{2n-2}{2n+1}\)

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

Để P nguyên thì \(\frac{3}{2n+1}\)nguyên

\(\Leftrightarrow3⋮\left(2n+1\right)\Leftrightarrow2n+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Lập bảng:

Vậy \(n\in\left\{-2;-1;0;1\right\}\)

#)Giải :

\(P=\frac{2n^2-n+2}{2n+1}=\frac{2n^2+n-2n-1+3}{2n+1}=\frac{n\left(2n+1\right)-\left(2n+1\right)+3}{2n+1}\)

\(=\frac{\left(2n+1\right)\left(n-1\right)+3}{2n+1}=n-1+\frac{3}{2n+1}\)

\(\Rightarrow2n+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Rightarrow\orbr{\begin{cases}2n+1=-3\\2n+1=1\end{cases}\Rightarrow\orbr{\begin{cases}n=-2\\n=-1\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}2n+1=1\\2n+1=3\end{cases}\Rightarrow\orbr{\begin{cases}n=0\\n=1\end{cases}}}\)

Vậy \(n\in\left\{-2;-1;0;1\right\}\)