phân tích
A = \(\sqrt{\left(\sqrt{3}+1\right)\left(4+2\sqrt{6}\right)}\)
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Phương pháp 2. Biến đổi về phương trình tích
a \(\sqrt{x^2-5x+6}+\sqrt{x+1}=\sqrt{x-2}+\sqrt{x^2-2x-3}\)
b \(2\sqrt[3]{\left(x+3\right)^2}-\sqrt[3]{\left(x-3\right)^2}=\sqrt[3]{x^2-9}\)
c \(\sqrt{2x+1}+3\sqrt{4x^2-2x+1}=3+\sqrt{8x^3+1}\)
d \(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)
a) ĐK: \(x\ge3\)
PT \(\Leftrightarrow\sqrt{\left(x-3\right)\left(x-2\right)}-\sqrt{x-2}+\sqrt{x+1}-\sqrt{\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-3}-1\right)+\sqrt{x+1}\left(1-\sqrt{x-3}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+1}\right)\left(\sqrt{x-3}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-2}=\sqrt{x+1}\\\sqrt{x-3}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=x+1\\x-3=1\end{matrix}\right.\) \(\Leftrightarrow x=4\) (Thỏa mãn)
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Rút gọn :
dfrac{sqrt{x+sqrt{4left(x-1right)}}-sqrt{x-sqrt{4left(x-1right)}}}{sqrt{x^2-4left(x-1right)}}.left(sqrt{x-1}-dfrac{1}{sqrt{x-1}}right)
b)left(sqrt{2}+1right)left(sqrt{3}+1right)left(sqrt{6}+1right)left(5-2sqrt{2}-sqrt{3}right)
c)left(sqrt{5}+1right)left(sqrt{7}+1right)left(sqrt{35}+1right)left(34-4sqrt{7}-6sqrt{5}right)
d) left(sqrt{7}+1right)left(2sqrt{2}-1right)left(2sqrt{14}-1right)left(55+12sqrt{2}-7sqrt{7}right)
e)left(3sqrt{2}+1right)left(2sqrt{3}+1right)left(6sqrt{6}+1...
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Rút gọn :
\(\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)\)
b)\(\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)\)
c)\(\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)\)
d) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
Tính:A2sqrt{left(-3right)^6}+2sqrt{left(-2right)^4}-4sqrt{left(-2right)^6}Bsqrt{left(sqrt{2}-2right)^2}+sqrt{left(sqrt{2}-3right)^2}Csqrt{left(3-sqrt{3}right)^2}-sqrt{left(1+sqrt{3}right)^2}Dsqrt{left(5+sqrt{6}right)^2}-sqrt{left(sqrt{6}-5right)^2}Esqrt{17^2-8^2}-sqrt{3^2+4^2}
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Tính:
\(A=2\sqrt{\left(-3\right)^6}+2\sqrt{\left(-2\right)^4}-4\sqrt{\left(-2\right)^6}\)
\(B=\sqrt{\left(\sqrt{2}-2\right)^2}+\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(C=\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(D=\sqrt{\left(5+\sqrt{6}\right)^2}-\sqrt{\left(\sqrt{6}-5\right)^2}\)
\(E=\sqrt{17^2-8^2}-\sqrt{3^2+4^2}\)
\(A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|\)
\(=54+8-32=30\)
\(B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}\)
\(=5-2\sqrt{2}\)
\(C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}\)
\(=2-2\sqrt{3}\)
\(D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}\)
\(=2\sqrt{6}\)
\(E=\sqrt{15^2}-\sqrt{5^2}=15-5=10\)
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`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`
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Tính:
1.sqrt{left(sqrt{3}+sqrt{2}right)^2} 4.sqrt{left(sqrt{3}right)^2+2.left(sqrt{3}right).left(1right)+left(1right)^2}
2.sqrt{left(sqrt{5}-sqrt{6}right)^2} 5.sqrt{left(sqrt{5}right)^2+2.left(sqrt{5}right).left(sqrt{3}right)+left(sqrt{3}right)^2}
3.sqrt{left(2sqrt{2}+sqrt{3}right)^2}...
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Tính:
1.\(\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\) 4.\(\sqrt{\left(\sqrt{3}\right)^2+2.\left(\sqrt{3}\right).\left(1\right)+\left(1\right)^2}\)
2.\(\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}\) 5.\(\sqrt{\left(\sqrt{5}\right)^2+2.\left(\sqrt{5}\right).\left(\sqrt{3}\right)+\left(\sqrt{3}\right)^2}\)
3.\(\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}\) 6.\(\sqrt{\left(\sqrt{6}\right)^2-2.\left(\sqrt{6}\right).\left(\sqrt{5}\right)+\left(\sqrt{5}\right)^2}\)
1> Rut gon
a)\(\sqrt{6-2\sqrt{2}+2\sqrt{3}-2\sqrt{6}}\)
b) \(\left(\sqrt{2}+1\right)\left(\left(\sqrt{2}\right)^2+1\right)\left(\left(\sqrt{2}\right)^4+1\right)\left(\left(\sqrt{2}\right)^8+1\right)\left(\left(\sqrt{2}\right)^{16}+1\right)\)
c)\(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
d) \(\sqrt{3-\frac{4\sqrt{5}}{3}}+\sqrt{3+\frac{4\sqrt{5}}{3}}\)
tính1.left(sqrt{15}-2sqrt{3}right)^2+12sqrt{5}2.3sqrt{2}left(4-sqrt{2}right)+3left(1-2sqrt{2}right)^23.dfrac{1}{2}left(sqrt{6}+sqrt{5}right)^2-dfrac{1}{4}sqrt{120}-sqrt{dfrac{15}{2}}4.left(sqrt{4-sqrt{7}}-sqrt{4+sqrt{7}}right)^25.left(sqrt{sqrt{14}+sqrt{5}}+sqrt{sqrt{14}-sqrt{5}}right)^26.left(sqrt{3}+1right)^3-left(sqrt{3}-1right)^37.left(sqrt{2}+1right)^3-left(sqrt{2}-1right)^38.sqrt{13-sqrt{160}}-sqrt{53+4sqrt{90}}9.sqrt{3-sqrt{5}}+sqrt{3+sqrt{5}}
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tính
1.\(\left(\sqrt{15}-2\sqrt{3}\right)^2+12\sqrt{5}\)
2.\(3\sqrt{2}\left(4-\sqrt{2}\right)+3\left(1-2\sqrt{2}\right)^2\)
3.\(\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
4.\(\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)^2\)
5.\(\left(\sqrt{\sqrt{14}+\sqrt{5}}+\sqrt{\sqrt{14}-\sqrt{5}}\right)^2\)
6.\(\left(\sqrt{3}+1\right)^3-\left(\sqrt{3}-1\right)^3\)
7.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
8.\(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
9.\(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)
Rút gọn :
a) left(sqrt{6}+sqrt{2}right).left(sqrt{3}-2right)left(sqrt{2+sqrt{3}}right)
b) sqrt{2}.left(sqrt{2-sqrt{3}}right).left(sqrt{3}+1right)
c) left(sqrt{10}-sqrt{6}right).left(sqrt{4-sqrt{15}}right)
d)left(sqrt{3}-sqrt{12}right).left(sqrt{5+2sqrt{6}}right)
e) sqrt{2-sqrt{3}}.left(sqrt{6}-sqrt{2}right).left(2+sqrt{3}right)
f) sqrt{4+sqrt{5sqrt{3}+5sqrt{48-10sqrt{7+4sqrt{3}}}}}
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Rút gọn :
a) \(\left(\sqrt{6}+\sqrt{2}\right).\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)\)
b) \(\sqrt{2}.\left(\sqrt{2-\sqrt{3}}\right).\left(\sqrt{3}+1\right)\)
c) \(\left(\sqrt{10}-\sqrt{6}\right).\left(\sqrt{4-\sqrt{15}}\right)\)
d)\(\left(\sqrt{3}-\sqrt{12}\right).\left(\sqrt{5+2\sqrt{6}}\right)\)
e) \(\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)\)
f) \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
a) Ta có: \(\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)\)
\(=\sqrt{2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{4+2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left|\sqrt{3}+1\right|\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)
\(=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)\)(Vì \(\sqrt{3}>1>0\))
\(=\left(4+2\sqrt{3}\right)\cdot\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(\sqrt{3}+2\right)\left(\sqrt{3}-2\right)\)
\(=2\cdot\left(3-4\right)\)
\(=-2\)
b) Ta có: \(\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}+1\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)(Vì \(\sqrt{3}>1\))
\(=3-1=2\)
c) Ta có: \(\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)\)
\(=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left|\sqrt{5}-\sqrt{3}\right|\cdot\left(\sqrt{5}-\sqrt{3}\right)\)
\(=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)(Vì \(\sqrt{5}>\sqrt{3}\))
\(=8-2\sqrt{15}\)
d) Ta có: \(\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)\)
\(=\sqrt{3}\cdot\left(1-2\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)
\(=-\sqrt{3}\cdot\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=-\sqrt{3}\cdot\left|\sqrt{3}+\sqrt{2}\right|\)
\(=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)\)(Vì \(\sqrt{3}>\sqrt{2}>0\))
\(=-3-\sqrt{6}\)
e) Ta có: \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left|\sqrt{3}-1\right|\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)\left(\sqrt{3}+2\right)\)(Vì \(\sqrt{3}>1\))
\(=\frac{\left(4-2\sqrt{3}\right)\left(4+2\sqrt{3}\right)}{2}\)
\(=\frac{16-12}{2}=\frac{4}{2}=2\)
f) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+2\cdot2\cdot\sqrt{3}+3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left|2+\sqrt{3}\right|}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)(Vì \(2>\sqrt{3}>0\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left|5-\sqrt{3}\right|}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)(Vì \(5>\sqrt{3}\))
\(=\sqrt{4+\sqrt{5\sqrt{3}+25-5\sqrt{3}}}\)
\(=\sqrt{4+\sqrt{25}}\)
\(=\sqrt{4+5}=\sqrt{9}=3\)
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Gấp lắm . Giúp mình cảm ơn ạBài 1 2sqrt{left(1+sqrt{3}right)^{ }3}-sqrt{left(2sqrt{3}-3right)^2}left(1+sqrt{3}-sqrt{5}right).left(1+sqrt{3}+sqrt{5}right)left(sqrt[]{dfrac{8}{3}}-sqrt{5}right)xsqrt{6}left(5+4sqrt{2}right).left(3+2sqrt{1}+sqrt{2}right).left(3-2sqrt{1}+2right)sqrt{3+2sqrt{2}}-sqrt{3-2sqrt{2}}
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Gấp lắm . Giúp mình cảm ơn ạ
Bài 1
\(2\sqrt{\left(1+\sqrt{3}\right)^{ }3}-\sqrt{\left(2\sqrt{3}-3\right)^2}\)
\(\left(1+\sqrt{3}-\sqrt{5}\right).\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(\left(\sqrt[]{\dfrac{8}{3}}-\sqrt{5}\right)x\sqrt{6}\)
\(\left(5+4\sqrt{2}\right).\left(3+2\sqrt{1}+\sqrt{2}\right).\left(3-2\sqrt{1}+2\right)\)
\(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
e) Ta có: \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2}+1-\sqrt{2}+1\)
=2
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19 tháng 8 2021 lúc 10:04
Tính
A= \(\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}+2\right)^2}\)
B=\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
C=\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
a: Ta có: \(A=\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=\sqrt{3}-1-2-\sqrt{3}\)
=-3
b: Ta có: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}-1\)
=1
c: Ta có: \(C=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=3-\sqrt{6}+2\sqrt{6}-3\)
\(=\sqrt{6}\)
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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...
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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 \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
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