câu a tớ giải được rồi, các bn giải câu b giùm mk
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Mình rút gọn như thế này đúng không nhỉ?Pleft(2-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{2x-sqrt{x}-3}+frac{sqrt{x}}{sqrt{x}+1}right)Pleft[frac{2left(2sqrt{x}-3right)}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right]:left[frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}+frac{sqrt{x}left(2sqrt{x}-3right)}{left(sqrt{x}+1right)left(2sqrt{x}-3right)}right]Pleft(frac{4sqrt{x}-6}{2sqrt{x}-3}-frac{sqrt{x}-1}{2sqrt{x}-3}right):left(frac{6sqrt{x}+1}{left(sqrt{x}+1right)left(2sqrt{x...
Đọc tiếp
Mình rút gọn như thế này đúng không nhỉ?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
rút gọn P=\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right):\left(\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{\sqrt{x}+3}-\frac{9-x}{1+\sqrt{x}-6}\right)\)
rút gọn:M=\(\left(\frac{\sqrt{x}-1}{\sqrt{x}+2}-\frac{\sqrt{x}+1}{2-\sqrt{x}}-\frac{2x-2\sqrt{x}}{x-4}\right)\cdot\frac{2\sqrt{x}-2}{3\sqrt{x}-6}\)
giải phương trình
a. \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=4-2x\)
b.\(\sqrt{x-2010}+\sqrt{y-2011}+\sqrt{x+2012}=\frac{1}{2}\left(x+y+z\right)-300\)
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-5}+2=0\)
\(2x-x^2+\sqrt{6x^2-12x+7}=0\)
\(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
\(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
giúp dùm đi mấy pạn
Tim x,y,z :1)left(2sqrt{x}-3right).left(2+sqrt{x}right)+602)sqrt{x-1+2sqrt{x-2}}+sqrt{x-1-2sqrt{x-2}}03)sqrt{x^2-4}-2sqrt{x-2}04)sqrt{x}+sqrt{y-1}+sqrt{z-2}frac{1}{2}.left(x+y+zright)5) xy xsqrt{y-1}+ysqrt{x-1}6)xsqrt{y-1}+2ysqrt{x-1}frac{3xy}{2}
Đọc tiếp
Tim x,y,z :
1)\(\left(2\sqrt{x}-3\right).\left(2+\sqrt{x}\right)+6=0\)
2)\(\sqrt{x-1+2\sqrt{x-2}}+\sqrt{x-1-2\sqrt{x-2}}=0\)
3)\(\sqrt{x^2-4}-2\sqrt{x-2}=0\)
4)\(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}.\left(x+y+z\right)\)
5) xy =\(x\sqrt{y-1}+y\sqrt{x-1}\)
6)\(x\sqrt{y-1}+2y\sqrt{x-1}=\frac{3xy}{2}\)
\(P=\frac{1}{\sqrt{x-1}-\sqrt{x}}+\frac{1}{\sqrt{x-1}+\sqrt{x}}+\frac{2\sqrt{x}-x}{\sqrt{x}-1}\left(x>1\right)\)
a) Rút gọn biểu thức
b) Tính giá trị khi \(x=\frac{53}{9-2\sqrt{7}}\)
\(Q=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
a. Rút gọn Q
b. Tìm x để Q = -1
Kamsamita. Saranghaeyo
cho biểu thức: P=\(\left[1-\frac{x-3\sqrt{x}}{x-9}\right]:\left[\frac{\sqrt{x}-3}{2-\sqrt{x}}+\frac{\sqrt{x}-2}{3+\sqrt{x}}-\frac{9x}{x+\sqrt{x}-6}\right]\) \(\left(x\le0;x\ne9;x\ne4\right)\)
a) Rút gọn P
b) Tìm giá trị của x để P=1