Nhìn không đủ chán rồi không dám động vào

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À do nãy máy lag sr :) Chứ bài đặt ẩn phụ mệt lắm :)

a) \(x+1=\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\)

<=> \(\left(x+1\right)^2=\left[\sqrt{2\left(x+1\right)+2\sqrt{2\left(x+1\right)+2\sqrt{4\left(x+1\right)}}}\right]^2\)

<=> \(x^2+2x+1=2x+2+2\sqrt{2x+2+4\sqrt{x+1}}\)

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

<=> \(x^2+1=2\sqrt{2x+2+4\sqrt{x+1}}+2\)

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

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

<=> \(\left(x^2-1\right)^2=\left(2\sqrt{2x+2+4\sqrt{x+1}}\right)^2\)

<=> \(x^4-2x^2+1=8x+8+16\sqrt{x+1}\)

<=> \(x^4-2x^2+1-8x=16\sqrt{x+1}+8\)

<=> \(x^4-2x^2-8x-7=16\sqrt{x+1}\)

<=> \(\left(x^4-2x^2-8x-7\right)^2=\left(16\sqrt{x+1}\right)^2\)

<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x+49=256x+256\)

<=> \(x^8-4x^6-16x^5-10x^4+32x^3+92x^2+112x-144x-207=0\)

<=> \(\left(x+1\right)\left(x-2\right)\left(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\right)=0\)

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vì: \(x^6+2x^5+3x^4-4x^3-9x^2+2x+69\ne0\)

=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Bạn xem lại đề câu b và c nhé !

a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)

\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)

\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ

\(\Rightarrow x\ge2\) thỏa mãn đề.

d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)

\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)

Pt tương đương :

\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )

e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)

\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)

Phương trình (1) tương đương :

\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)

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

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

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )

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

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

\(\Rightarrow\sqrt{\left(x+\frac{1}{2}\right)^2}=x+\frac{1}{2}=x^3+\frac{x^2}{2}+x+\frac{1}{2}\)

\(\Rightarrow x^3+\frac{x^2}{2}+x+\frac{1}{2}-x-\frac{1}{2}=x^3+\frac{x^2}{2}=0\Rightarrow\frac{2x^3+x^2}{2}=0\)

\(\Rightarrow2x^3+x^2=0\Rightarrow x^2\left(2x+1\right)=0\Rightarrow\hept{\begin{cases}x^2=0\Rightarrow x=0\\2x+1=0\Rightarrow2x=-1\Rightarrow x=-\frac{1}{2}\end{cases}}\)

vậy x=0 và x=-1/2

\(S=\frac{-1+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{99}+\sqrt{100}}{100-99}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{99}+\sqrt{100}\)

\(=-1+\sqrt{100}\)

\(\hept{\begin{cases}a=\left(x^2-x+1\right)^2\\b=x^2\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=b\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x^2-x+1\right)^2=1\\\left(x^2-x+1\right)^2=x^2\end{cases}}\)(easy)

\(\sqrt{x^2-\frac{1}{4}+\sqrt{x^2+x+\frac{1}{4}}}=\frac{1}{2}\left(2x^3+x^2+2x+1\right)\) (*) (ĐKXĐ: \(\forall x\in R\))

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

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

+) Xét \(x+\frac{1}{2}\ge0\Leftrightarrow x\ge-\frac{1}{2}\). Khi đó pt (*) trở thành:

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

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

\(\Leftrightarrow x+\frac{1}{2}=\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)\) (Do \(x\ge\frac{1}{2}\))

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

\(\Leftrightarrow x^2\left(2x+1\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\) (t/m ĐKXĐ)

+) Xét \(x+\frac{1}{2}< 0\Leftrightarrow x< -\frac{1}{2}\). Khi đó: \(2x+1< 0\)

Ta thấy: \(2x+1< 0;x^2+1>0;\frac{1}{2}>0\Rightarrow\frac{1}{2}\left(2x+1\right)\left(x^2+1\right)< 0\)

Mà \(\sqrt{x^2-\frac{1}{4}+\left|x+\frac{1}{2}\right|}\ge0\) nên Vô lí ---> Loại TH này.

Vậy tập nghiệm của pt (*) là \(S=\left\{0;-\frac{1}{2}\right\}.\)

rthgsdgdh olweikehgf

Sorry nha nhưng em mới học lớp 7 thôi à ~~

6/ Đặt \(\hept{\begin{cases}\sqrt[4]{x}=a\\\sqrt[4]{2-x}=b\end{cases}}\)

\(\Rightarrow b^4+a^4=2\)

Từ đó ta có: a + b = 2

Ta có: \(a^4+b^2\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(a+b\right)^4}{8}=\frac{16}{8}=2\)

Dấu = xảy ra khi a = b = 1

=> x = 1