\(\sqrt[3]{x^2}+\sqrt[3]{x+1}=\sqrt[3]{x}+\sqrt[3]{x^2+x}\)

\(\Leftrightarrow\sqrt[3]{x^2}-1+\sqrt[3]{x+1}-\sqrt[3]{2}=\sqrt[3]{x}-1+\sqrt[3]{x^2+x}-\sqrt[3]{2}\)

\(\Leftrightarrow\frac{x^2-1}{\sqrt[3]{x^2}^2+\sqrt[3]{x^2}+1}+\frac{x+1-2}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}\sqrt[3]{2}+\sqrt[3]{2}^2}=\frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}+\frac{x^2+x-2}{\sqrt[3]{x^2+x}^2+\sqrt[3]{x^2+x}\sqrt[3]{2}+\sqrt[3]{2}^2}\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+1\right)}{\sqrt[3]{x^2}^2+\sqrt[3]{x^2}+1}+\frac{x-1}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}\sqrt[3]{2}+\sqrt[3]{2}^2}-\frac{x-1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}-\frac{\left(x-1\right)\left(x+2\right)}{\sqrt[3]{x^2+x}^2+\sqrt[3]{x^2+x}\sqrt[3]{2}+\sqrt[3]{2}^2}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+1}{\sqrt[3]{x^2}^2+\sqrt[3]{x^2}+1}+\frac{1}{\sqrt[3]{x+1}^2+\sqrt[3]{x+1}\sqrt[3]{2}+\sqrt[3]{2}^2}-\frac{1}{\sqrt[3]{x}^2+\sqrt[3]{x}+1}-\frac{x+2}{\sqrt[3]{x^2+x}^2+\sqrt[3]{x^2+x}\sqrt[3]{2}+\sqrt[3]{2}^2}\right)=0\)

Suy ra x=1. pt kia chịu :v nghiệm lẻ quá

Thắng Nguyễn đúng là thánh troll

đặt \(\sqrt[3]{x}=a;\sqrt[3]{x+1}=b\)

pt trở thành:

a²+b=a+ab

<=>a(a-1)-b(a-1)=0

<=>(a-b)(a-1)=0

từ đó thay vào rồi giải tìm x

a, ĐK \(\hept{\begin{cases}x>0\\x\ne1\end{cases}}\)

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

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

Ta thấy \(P=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}-1}>0\forall x>0,x\ne1\)

b, P=\(\frac{x+2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\frac{2}{2+\sqrt{3}}+2\sqrt{\frac{2}{2+\sqrt{3}}}+1}{\sqrt{\frac{2}{2+\sqrt{3}}}-1}\)

=\(\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\sqrt{\left(\frac{2}{\left(\sqrt{3}+1\right)^2}\right)}+1}{\sqrt{\left(\frac{2}{2+\sqrt{3}}\right)^2}-1}=\frac{\frac{4}{\left(\sqrt{3}+1\right)^2}+2.\frac{2}{\sqrt{3}+1}+1}{\frac{2}{\sqrt{3}+1}-1}\)

\(=\frac{12+6\sqrt{3}}{1-3}=-6-3\sqrt{3}\)

cậu ơi câu c đâu ạ??

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

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

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

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

Dễ thấy: \(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}< 0\)

\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)

a. ĐK \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

b. M =\(\frac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+2\sqrt{x}+1-x+2\sqrt{x}-1-5\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

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

c. \(M=\frac{-1}{\sqrt{x}+1}\ge-1\)

Vậy Min M =-1 khi x=0

thanks nha bạn

a,=\(4x-\sqrt{\left(x-2\right)^2}\)

=\(4x-x+2\)

=3x+2

b,\(3x+\sqrt{\left(x+3\right)^2}\)

=\(3x+x+3\)

=4x+3

a,bạn viết thiếu đầu bài

b,<=>3x-2=4

<=>3x=6

<=>x=2

vậy...........................

c,=>\(5\left(2\sqrt{x}-19\right)=4-\sqrt{x}\)ĐKXĐ x>=0 x khác 16

<=>\(10\sqrt{x}-95-4+\sqrt{x}=0\)

<=>\(11\sqrt{x}-99=0\)

<=>\(11\sqrt{x}=99\)

<=>\(\sqrt{x}=9< =>x=81\)

vậy.............

k mk nha

#quynh tong ngoc ơi, câu a đề bài là vậy rồi nhé >< Mình viết đúng đấy bạn ạ

a . ĐKXĐ \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

P=\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}+\frac{8\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x-\sqrt{x}-2\sqrt{x}-2+8\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+5\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{\sqrt{x}+6}{\sqrt{x}+1}\)

b. P =4\(\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}+1}=4\Leftrightarrow3\sqrt{x}=2\Leftrightarrow x=\frac{4}{9}\)

c. \(P>7\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}+1}-7>0\Leftrightarrow\frac{-\sqrt{x}-1}{\sqrt{x}+1}>0\)

\(\Leftrightarrow\sqrt{x}< -1\)vô nghiệm

Vậy không tồn tại x để P >7

d. \(P=\frac{\sqrt{x}+6}{\sqrt{x}+1}=1+\frac{5}{\sqrt{x}+1}\)

Ta thấy \(\sqrt{x}+1\ge1\Rightarrow\frac{5}{\sqrt{x}+1}\le5\Rightarrow P\le6\)

Vậy Max P =6.Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)