Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

\(\frac{x-37}{1943}+\frac{x-35}{1945}+\frac{x-33}{1947}+\frac{x-31}{1949}=\frac{x-1943}{37}+\frac{x-1945}{35}+\frac{x-1947}{33}+\frac{x-1949}{31}\)

\(\Leftrightarrow\frac{x-37}{1943}-1+\frac{x-35}{1945}-1+\frac{x-33}{1947}-1+\frac{x-31}{1949}-1=\frac{x-1943}{37}-1+\frac{x-1945}{35}-1+\frac{x-1947}{33}-1+\frac{x-1949}{31}-1\)

\(\Leftrightarrow\frac{x-1980}{1943}+\frac{x-1980}{1945}+\frac{x-1980}{1947}+\frac{x-1980}{1949}=\frac{x-1980}{37}+\frac{x-1980}{35}+\frac{x-1980}{33}+\frac{x-1980}{31}\)

\(\Leftrightarrow x-1980=0\)

\(\Leftrightarrow x=1980\)

\(\left(x+3\right)^2-\left(2x+6\right)\left(1-3x\right)+\left(3x+1\right)^2\)

\(=x^2+6x+9-\left(2x-6x^2+6-18x\right)+9x^2+6x+1\)

\(=10x^2+12x+10+6x^2+16x-6=16x^2+28x+4\)

\(=4\left(4x^2+7x+1\right)\)

\(P=4x^2+2y^2-4xy+4x-4y+2021\)

\(=4x^2+y^2+1-4xy+4x-2y+y^2-2y+1+2019\)

\(=\left(2x-y+1\right)^2+\left(y-1\right)^2+2019\ge2019\)

Dấu \(=\)khi \(\hept{\begin{cases}2x-y+1=0\\y-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=1\end{cases}}\).

sau bạn đăng tách ra cho mn cùng giúp nhé 

a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)

b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)

c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)

d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)

e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)

f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)

\(\left(x-1\right)\left(x-2\right)\left(x+4\right)\left(x+5\right)=122\)

\(\Leftrightarrow\left(x^2+3x-4\right)\left(x^2+3x-10\right)=122\)

Đặt \(x^2+3x-4=t\)

\(\Leftrightarrow t\left(t-6\right)=122\Leftrightarrow t^2-6t-122=0\Leftrightarrow t=3\pm\sqrt{131}\)

đề có lỗi ko bạn ? 

a) Xét tứ giác \(ABCD\)có:

\(A+B+C+D=360\)độ

\(\Rightarrow\)\(\frac{A}{1}=\frac{B}{2}=\frac{C}{3}=\frac{D}{4}=\frac{A+B+C+D}{1+2+3+4}\)

\(=\)\(\frac{360}{10}=36\)

\(\Rightarrow\)\(A=36;B=72;C=108;D=144\)độ

b) Ta có: \(A+D=180\)độ

Mà 2 góc này ở vị trí trong cùng phía bù nhau

\(\Rightarrow\)\(AB//CD\)

c) Vì      \(AB//CD\left(cmt\right)\)

\(\Rightarrow\)\(\hept{\begin{cases}EDC=36\\ECD=72\end{cases}}\)độ ( 2 góc đồng vi )

Xét \(\Delta EDC\)có:

\(EDC+ECD+CED=180\)Độ

\(\Rightarrow\)\(CED=180\) \(-\left(EDC+ECD\right)=180\)\(-\left(36+72\right)\)\(=72\)ĐỘ

cần gấp mik k cho