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

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

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

\(\Leftrightarrow\sqrt{1-x}=0\Leftrightarrow x=1.\)

toán lớp 9 thì ai mà biết chỉ lớp 5 thôi

đáp án là : 0 bít !

sống bớt xàm đi bạn trẻ

ặc vô nghiệm nữa rồi mong ko sai đề tiếp :V

ĐKXĐ: x lớn hơn hoặc bằng -1 và x nhỏ hơn hoặc bằng 1.

\(4+2\sqrt{1-x}=-3x+5\sqrt{x+1}+\sqrt{1-x^2}\)

\(\Leftrightarrow4+2\left(\sqrt{1-x}-1\right)+2=-3x+5\left(\sqrt{x+1}-1\right)+\left(\sqrt{1-x^2}-1\right)+5+1\)

\(\frac{-2x}{\sqrt{1-x}+1}=-3x+\frac{5x}{\sqrt{x+1}+1}-\frac{x^2}{\sqrt{1-x^2}+1}\Leftrightarrow x\left(\frac{x}{\sqrt{1-x^2}+1}-\frac{5}{\sqrt{x+1}+1}-\frac{2}{\sqrt{1-x}+1}+3\right)=0\)

\(\Leftrightarrow x=0.\)

\(Pt\Leftrightarrow3\left(x+1\right)+2\sqrt{1-x}+1=5\sqrt{x+1}+\sqrt{1-x^2}\)

đặt \(\sqrt{x+1}=a,\sqrt{1-x}=b\)

\(\Leftrightarrow3a^2+2b+1=a\left(5+b\right)\)

\(\Leftrightarrow3a^2-\left(5+b\right)a+2b+1=0\)

\(\Delta=b^2-4ac=\left(-b-5\right)^2-4.3.\left(2b+1\right)\)

\(=b^2+10b+25-24b-12\)

\(=b^2-14b+13\)

\(TH1:\Rightarrow a=\frac{5+b+\sqrt{b^2-14b+13}}{6}\)

\(\Rightarrow6a-5-b=\sqrt{b^2-14b+13}\)

\(\Rightarrow6\sqrt{1+x}-5-\sqrt{1-x}=\sqrt{1-x-14\sqrt{1-x}+13}\)

\(\hept{\begin{cases}x=0\left(nhan\right)\\x=......\left(loai\right)\end{cases}}\)

TH2:\(a=\frac{5+b-\sqrt{b^2-14b+13}}{6}\)

\(.............................................\)

cách này hơi dài.

Ta có \(\left(a+b+c+1\right)^2=\left(\left(a+1\right)+b+c\right)^2\)

\(=a^2+2a+1+b^2+c^2+2b+2c+2ab+2bc+2ac\left(f\right)\)

Từ \(\left(f\right)\Rightarrow2a+2b+2c+2ab+2bc+2ac\ge3\left(a^2+b^2+c^2\right)\)

Mà \(a,b,c\in\left[0;1\right]\)nên \(a\ge a^2,b\ge b^2,c\ge c^2\left(g\right)\)

Từ \(\left(f\right)vs\left(g\right)\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\)

Áp dụng bất đẳng thức cô si ta có:

\(2ab+2bc+2ac\ge3\sqrt[3]{8\left(abc\right)^2}=6\sqrt[3]{\left(abc\right)^2}\)

\(a^2+b^2+c^2\ge3\sqrt[3]{\left(abc\right)^2}\)

\(\Rightarrow2ab+2bc+2ac\ge a^2+b^2+c^2\Rightarrowđpcm\).Dấu bằng tự tìm nha.

Đặt \(a=\sqrt{2-x^2};b=\sqrt{2-\frac{1}{x^2}};c=x+\frac{1}{x}\)

xet x<0 vt < 2 căn 2<3, vt >4=>loại=>x>0=>c>=2;

ta có a+b=4-c;

a^2+b^2=4-x^2-1/x^2=6-c^2;

\(=>\hept{\begin{cases}2a+2b=8-2c\left(2\right)\\a^2+b^2=6-c^2\left(1\right)\end{cases}}\)

trừ 1 cho 2=>a^2-2a+b^2-2b=-c^2-2-2c=>a^2-2b+1+b^2-2b+1=-c^2+2c-1+1

=>\(\left(a-1\right)^2+\left(b-1\right)^2=-\left(c-1\right)^2+1\)

\(< =>\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=1\)

ta lại có (a-1)^2>=0;(b-1)^2>=0;(c-1)^2>=(2-1)^2=1=>Vế trái>=1=Vế phải, dấu bằng xảy ra<=>

\(\hept{\begin{cases}a=1\\b=1\\c=2\end{cases}< =>x=1}\)

Bạn tham khảo nhé:Điều kiện bạn tự tìm nhé

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

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

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\Leftrightarrow x=1\left(N\right)\\\frac{1}{\sqrt{2-x^2}-x+2}+\frac{1}{x\sqrt{2x^2-1}-x+2x^2}=0\left(1\right)\end{cases}}\)

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

Nhân 2 vào ta có:

\(\Leftrightarrow2x\sqrt{2x^2-1}-4x+4x^2+4+2\sqrt{2-x^2}=0\)

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

Vậy phương trình có 1 nghiệm duy nhất là \(x=1\)

Bổ sung cách độc lạ hơn nè mình vừa nghĩ ra:

Chuyển vế:

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

Ap dụng BĐT a+b<=\(\sqrt{2\left(a^2+b^2\right)}\)

Dấu = khi a=b

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

=2+2=4=VP. Dấu = xảy ra khi \(\hept{\begin{cases}\sqrt{2-x^2}=x\\\sqrt{2-\frac{1}{x^2}}=\frac{1}{x}\end{cases}< =>x=1}\)

dùng bđt xem sao

\(\Leftrightarrow\left(3-x\right)\sqrt{x-1}+\sqrt{5-2x}=\sqrt{\left[\left(x-3\right)^2+1\right]\left(4-x\right)}\)

đặt 3-x=a;\(\sqrt{x-1}=b;\sqrt{5-2x}=c\Rightarrow b^2+c^2=4-x\)

\(\Leftrightarrow ab+c=\sqrt{\left(a^2+1\right)\left(b^2+c^2\right)}\)

<=>a²b²+2abc+c²=a²b²+b²+a²c²+c²

<=>b²-2abc+a²c²=0

<=>(b-ac)²=0

<=>b=ac

đến đây thì dễ rồi