a, Ta có :
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x+\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}-2}\)sử dụng tam thức bậc 2 khai triển biểu thức trên tử nhé
\(=\frac{\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=\frac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\left(\sqrt{x}\right)^3-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}=x-1\)
b, Ta có : \(P=Q\)hay \(2\sqrt{x}+1=x-1\Leftrightarrow-x+2\sqrt{x}+2=0\)
\(\Leftrightarrow x-2\sqrt{x}-2=0\Leftrightarrow x-2\sqrt{x}+1-3=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2-3=0\Leftrightarrow\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
TH1 : \(\sqrt{x}=1+\sqrt{3}\Leftrightarrow x=\left(1+\sqrt{3}\right)^2=1+2\sqrt{3}+3=4+2\sqrt{3}\)
TH2 : \(\sqrt{x}=1-\sqrt{3}\Leftrightarrow x=\left(1-\sqrt{3}\right)^2=1-2\sqrt{3}+3=4-2\sqrt{3}\)
Vậy \(x=4+2\sqrt{3};x=4-2\sqrt{3}\)thì P = Q
んuリ イ giải pt vô tỉ không xét ĐK là tai hại :))
\(P=\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\frac{2x-4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-2\right)+\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=\frac{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}=2\sqrt{x}+1\)
\(Q=\frac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}=\frac{\left(x\sqrt{x}-\sqrt{x}\right)+\left(2x-2\right)}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(x-1\right)+2\left(x-1\right)}{\sqrt{x}+2}=\frac{\left(x-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=x-1\)
Để P = Q thì \(2\sqrt{x}+1=x-1\)( x ≥ 1 ; x ≠ 4 )
<=> \(x-2\sqrt{x}-2=0\)
<=> \(\left(\sqrt{x}-1\right)^2-3=0\)
<=> \(\left(\sqrt{x}-1-\sqrt{3}\right)\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
<=> \(\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}\Rightarrow\orbr{\begin{cases}x=4+2\sqrt{3}\left(tm\right)\\x=4-2\sqrt{3}\left(ktm\right)\end{cases}}\)
Vậy với \(x=4+2\sqrt{3}\)thì P = Q
a)
+) P=2x−3√x−2√x−2
=(2√x+1)(√x−2)√x−2
=2√x+1.
+) Q=√x3−√x+2x−2√x+2
=(√x+2)(x−1)√x+2
=x−1.
b) Để thì 2√x+1=x−1
Coi phương trình là phương trình bậc 2 của , chú ý chọn nghiệm dương của phương trình.
Đáp số: x=4+2√3 thì .
Câu hỏi thuộc chủ đề: Rút gọn biểu thức chứa căn bậc hai
o l m . v n
a, P= \(\sqrt{x}+\dfrac{1}{2}\)
Q= \(x-1\)
b, \(x= \dfrac{4+ \sqrt{7}}{2} hoặc x= \dfrac{4-\sqrt{7}}{2}\)
a, P=2\(\sqrt{x}+1\) ; Q=x-1 b, Để P=Q khi x=16
a/ P=\(\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}=\dfrac{\left(\sqrt{x}-2\right).\left(2\sqrt{x}+1\right)}{\sqrt{x}-2}\)=\(2\sqrt{x}+1\)
Q=\(\dfrac{\sqrt{x^3}+2x-\sqrt{x}-1}{\sqrt{x}+2}=\dfrac{\left(x-1\right).\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)=x-1
Vậy P=\(2\sqrt{x}+1\);Q=x-1 với x≥0 ; x≠4
b/ Để P=Q⇒\(2\sqrt{x}+1\)= x-1
⇒x \(-2\sqrt{x}-2\)=0 ⇔x= 4-\(2\sqrt{3}\)
Vậy x=4-\(2\sqrt{3}\) thì P=Q
a) P= \(2\sqrt{x}+1\)
Q= x-1
b) x=4+2\(\sqrt{3}\)
a) Với x >0 ; x≠4 ,ta có:
P=\(\dfrac{2x-4\sqrt{x}+\sqrt{x}+2}{\sqrt{x}-2}\)
=\(\dfrac{(\sqrt{x}-2)\times(2\sqrt{x}+1)}{\sqrt{x}-2}\)
P=\(2\sqrt{x}+1\)
Q=\(\dfrac{x\sqrt{x}-\sqrt{x}+2(x-1)}{\sqrt{x}+2}\)
=\(\dfrac{(\sqrt{x}+2)\times(x-1)}{\sqrt{x}+2}\)
Q=x-1
Vậy với x≥0;x≠4 , ta có P=\(2\sqrt{x}+1\) ; Q=x-1
b) Để P=Q
⇔\(2\sqrt{x}+1=x-1\)
⇔\(x-2\sqrt{x}-2=0\)⇔\((\sqrt{x}-1-\sqrt{3})\times(\sqrt{x}-1+\sqrt{3})=0\)
⇔\(\left[{}\begin{matrix}\sqrt{x}=1+\sqrt{3}\\\sqrt{x}=1-\sqrt{3}(loại\end{matrix}\right.\Leftrightarrow x=4+2\sqrt{3}\)( thỏa mãn x ≥0;x≠4)
Vậy với x =\(4+2\sqrt{3}\) thì P=Q
a, P = \(\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
P = \(2\sqrt{x}+1\)
Q = \(\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
Q = \(x-1\)
b Ta có để P=Q thì :
\(\Rightarrow2\sqrt{x}+1=x-1\)
\(\Rightarrow x=4+2\sqrt{3}\)
a,\(P=\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)
P=\(\dfrac{\left(2\sqrt{x}+1\right).\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
P=\(2\sqrt{x}+1\)
Q=\(\dfrac{\left(\sqrt{x}+2\right).\left(x-1\right)}{\sqrt{x}+2}\)
Q=x-1
b,P=Q
\(< =>2\sqrt{x}+1=x-1\)
\(< =>-x+2\sqrt{x}+2=0\)
\(< =>\left(\sqrt{x}-1\right).\left(\sqrt{x}-1\right)-3=0\)
\(< =>\left(\sqrt{x}-1-\sqrt{3}\right).\left(\sqrt{x}-1+\sqrt{3}\right)=0\)
\(< =>x=4+2\sqrt{3}\)
Vậy x = \(4+2\sqrt{3}\)thì P=Q
a)
+) P=\(\dfrac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)
=\(\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
=2\(\sqrt{x}+1\)
+) Q=\(\dfrac{\sqrt{x^3}-\sqrt{x}+2x-2}{\sqrt{x}+2}\)
=\(\dfrac{\left(\sqrt{x}+2\right)\left(x-1\right)}{\sqrt{x}+2}\)
=x-1
Để P=Q thi \(2\sqrt{x}+1\)= x-1
Đáp số: x=4+\(2\sqrt{3}\)thì P=Q
a) P=\(2\sqrt{x}+1\)
Q=x-1
a, P=2\(\sqrt{x}\)+ 1; Q = x- 1 với x >= 0; x khác 4
b, x = \(4+2\sqrt{3}\)thì P =Q