\(\sqrt{\left(x+1\right)\left(2-x\right)}=1+2x-2x^2\)
Những câu hỏi liên quan
\(\frac{\left(2x\sqrt{2}+2\sqrt{2x}\right)\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right)}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x}+1\right).\left(-2\sqrt{x}-2\right)}\)
=\(\frac{2\sqrt{2x}\left(\sqrt{x}+1\right)}{-2\left(\sqrt{x}+1\right)}\)
=\(-\sqrt{2x}\)
Pleft(frac{sqrt{x}+1}{sqrt{2x}+1}+frac{sqrt{2x}+sqrt{x}}{sqrt{2x}-1}-1right):left(1+frac{sqrt{x}+1}{sqrt{2x}+1}-frac{sqrt{2x}-sqrt{x}}{sqrt{2x}-1}right)frac{left(sqrt{x}+1right)left(sqrt{2x}-1right)}{left(sqrt{2x}+1right)left(sqrt{2x}-1right)}+frac{left(sqrt{2x}+sqrt{x}right)left(sqrt{2x}+1right)}{MTC}-frac{2x-1}{MTC}frac{xsqrt{2}-sqrt{x}+sqrt{2x}-1+2x+sqrt{2x}+xsqrt{2}+sqrt{x}-2x+1}{MTC}frac{2xsqrt{2}+2sqrt{2x}}{MTC}
Đọc tiếp
P=\(\left(\frac{\sqrt{x}+1}{\sqrt{2x}+1}+\frac{\sqrt{2x}+\sqrt{x}}{\sqrt{2x}-1}-1\right):\left(1+\frac{\sqrt{x}+1}{\sqrt{2x}+1}-\frac{\sqrt{2x}-\sqrt{x}}{\sqrt{2x}-1}\right)\)
=\(\frac{\left(\sqrt{x}+1\right)\left(\sqrt{2x}-1\right)}{\left(\sqrt{2x}+1\right)\left(\sqrt{2x}-1\right)}+\frac{\left(\sqrt{2x}+\sqrt{x}\right)\left(\sqrt{2x}+1\right)}{MTC}-\frac{2x-1}{MTC}\)
=\(\frac{x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1+2x+\sqrt{2x}+x\sqrt{2}+\sqrt{x}-2x+1}{MTC}\)
=\(\frac{2x\sqrt{2}+2\sqrt{2x}}{MTC}\)
\(\frac{2x-1}{MTC}+\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{MTC}-\frac{\left(\sqrt{2x}+\sqrt{x}\left(\sqrt{2x}+1\right)\right)}{MTC}\)
=\(\frac{2x-1+x\sqrt{2}-\sqrt{x}+\sqrt{2x}-1-2x-\sqrt{2x}-x\sqrt{2}-\sqrt{x}}{MTC}\)
=\(\frac{-2\sqrt{x}-2}{\left(\sqrt{2x}-1\right)\left(\sqrt{2x+1}\right)}\)
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
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?
\(P=\left(2-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{2x-\sqrt{x}-3}+\frac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(P=\left[\frac{2\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right]:\left[\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right]\)
\(P=\left(\frac{4\sqrt{x}-6}{2\sqrt{x}-3}-\frac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}+\frac{2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\left(\frac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}\right):\left(\frac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\right)\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}:\frac{2x+3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}\)
\(P=\frac{3\sqrt{x}-5}{2\sqrt{x}-3}.\frac{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-3\right)}{2x+3\sqrt{x}+1}\)
\(P=\left(3\sqrt{x}-5\right).\frac{\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x+3\sqrt{x}-5\sqrt{x}-5}{2x+3\sqrt{x}+1}\)
\(P=\frac{3x-5\sqrt{x}-5}{2x+1}\)
từ dòng cuối là sai rồi bạn à
Bạn bỏ dòng cuối đi còn lại đúng rồi
Ở tử đặt nhân tử chung căn x chung rồi lại đặt căn x +1 chung
Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra
rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
Đúng 0
Bình luận (0)
14 tháng 1 2021 lúc 12:59
Giải phương trình 1, \(x^2+9x+7=\left(2x+1\right)\sqrt{2x^2+4x+5}\)
2, GPT \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
3. GHPT \(\left\{{}\begin{matrix}x^2-2y-1=2\sqrt{5y+8}+\sqrt{7x-1}\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
1.
\(\Leftrightarrow\left(2x+1\right)\sqrt{2x^2+4x+5}-\left(2x+1\right)\left(x+3\right)+x^2-2x-4=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\sqrt{2x^2+4x+5}-\left(x+3\right)\right)+x^2-2x-4=0\)
\(\Leftrightarrow\dfrac{\left(2x+1\right)\left(x^2-2x-4\right)}{\sqrt{2x^2+4x+5}+x+3}+x^2-2x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\\dfrac{2x+1}{\sqrt{2x^2+4x+5}+x+3}+1=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x+1+\sqrt{2x^2+4x+5}+x+3=0\)
\(\Leftrightarrow\sqrt{2x^2+4x+5}=-3x-4\) \(\left(x\le-\dfrac{4}{3}\right)\)
\(\Leftrightarrow2x^2+4x+5=9x^2+24x+16\)
\(\Leftrightarrow7x^2+20x+11=0\)
Đúng 3
Bình luận (1)
2.
ĐKXĐ: ...
\(\Leftrightarrow2x\sqrt{2x+7}+7\sqrt{2x+7}=x^2+2x+7+7x\)
\(\Leftrightarrow\left(x^2-2x\sqrt{2x+7}+2x+7\right)+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)^2+7\left(x-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{2x+7}\right)\left(x+7-\sqrt{2x+7}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2x+7}\\x+7=\sqrt{2x+7}\end{matrix}\right.\)
\(\Leftrightarrow...\)
Đúng 3
Bình luận (1)
3.
ĐKXĐ: ...
Từ pt dưới:
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-y^3+3x-3y=3x^2+3y^2+1+1\)
\(\Leftrightarrow x^3-3x^2+3x-1=y^3+3y^2+3y+1\)
\(\Leftrightarrow\left(x-1\right)^3=\left(y+1\right)^3\)
\(\Leftrightarrow y=x-2\)
Thế vào pt trên:
\(x^2-2x+3=2\sqrt{5x-2}+\sqrt{7x-1}\)
\(\Leftrightarrow x^2-5x+2+2\left(x-\sqrt{5x-2}\right)+\left(x+1-\sqrt{7x-1}\right)=0\)
\(\Leftrightarrow x^2-5x+2+\dfrac{2\left(x^2-5x+2\right)}{x+\sqrt{5x-2}}+\dfrac{x^2-5x+2}{x+1+\sqrt{7x-1}}=0\)
\(\Leftrightarrow x^2-5x+2=0\)
Đúng 2
Bình luận (2)
giải phương trình :
a, \(\left(x+9\right)\left(2-\sqrt{9+2x}\right)^2=2x^2\)
b,\(\left(2x+10\right)\left(1-\sqrt{3+2x}\right)^2=4\left(x+1\right)^2\)
a. Đề bài sai, phương trình không giải được
b.
ĐKXĐ: \(x\ge-\dfrac{2}{3}\)
\(\left(2x+10\right)\left(\dfrac{1-\left(3+2x\right)}{1+\sqrt{3+2x}}\right)^2=4\left(x+1\right)^2\)
\(\Leftrightarrow\dfrac{\left(2x+10\right)4.\left(x+1\right)^2}{\left(1+\sqrt{3+2x}\right)^2}=4\left(x+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}4\left(x+1\right)^2=0\Rightarrow x=-1\\2x+10=\left(1+\sqrt{3+2x}\right)^2\left(1\right)\end{matrix}\right.\)
Xét (1)
\(\Leftrightarrow2x+10=2x+4+2\sqrt{2x+3}\)
\(\Leftrightarrow\sqrt{2x+3}=3\)
\(\Leftrightarrow x=3\)
Đúng 1
Bình luận (2)
Rút gọn:
\(A=1-\left[\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}+\dfrac{2x-1+\sqrt{x}}{1-x}\right]\cdot\left[\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right]\)
\(B=\left[1:\frac{2x-1}{x-x^2}\right]\cdot\left[\frac{2x^3+x^2-x}{x^3-1}-2-\frac{1}{x-1}\right]\)
\(\left|2x+3\right|+\left|2x-1\right|=\dfrac{8}{3.\left(x+1\right)^2+2}\)\(\sqrt{ }\)\(\left|2x+3\right|+\left|2x-1\right|\)=\(\dfrac{8}{3.\left(x+1\right)^2+2}\)