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b) Ta có: \(4x^2+x-5=0\)
\(\Leftrightarrow4x^2-4x+5x-5=0\)
\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)
Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:
\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)
Vậy: Khi \(4x^2+x-5=0\) thì B=0
Đề bài sai nhé, từ giả thiết chỉ xác định được \(x+y=0\Rightarrow y=-x\)
\(\Rightarrow A=4x^2-x^2+x^2+15=4x^2+15\) ko rút gọn được
Nguyễn Việt Lâm Giáo viên, bn có thể sửa đề bài cho mk được không ạ??? Cám ơn bn nhiều lắm lắm!!!
Đặt \(a=\sqrt[3]{9+4\sqrt{5}},b=\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow\hept{\begin{cases}a^3+b^3=18\\ab=1\end{cases};a+b=x}\)
Ta có: \(x=a+b\Leftrightarrow x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)\(\Rightarrow x^3=18+3x\Leftrightarrow x^3-3x=18\)(1)
Tương tự: Đặt \(c=\sqrt[3]{3+2\sqrt{2}},d=\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow\hept{\begin{cases}c^3+d^3=6\\cd=1\end{cases};c+d=y}\)
Ta có: \(y=c+d\Leftrightarrow y^3=\left(c+d\right)^3=c^3+d^3+3cd\left(c+d\right)\)\(\Rightarrow y^3=6+3y\)
\(\Leftrightarrow y^3-3y=6\)(2)
Từ (1) và (2) suy ra \(A=x^3-3x+y^3-3y+2020=18+6+2020=2048\)
a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left|x-2y\right|\)
TH1: \(x-2y--\left(x-2y\right)\)
\(=x-2y+x-2y\)
\(=2x-4y\)
TH2: \(x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
b) \(x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+\left|x^2-4\right|\)
TH1:
\(x^2+-\left(x^2-4\right)\)
\(=x^2-x^2+4\)
\(=4\)
TH2:
\(x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)
\(=2x-1-\sqrt{x-5}\)
d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))
\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)
\(=\sqrt{x^2-2}\)
e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+1\)
TH1:
\(x^2-4+1\)
\(=x^2-3\)
TH2:
\(-\left(x^2-4\right)+1\)
\(=-x^2+4+1\)
\(=-x^2+5\)
a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)
=x-2y-|x-2y|
Khi x>=2y thì A=x-2y-x+2y=0
Khi x<2y thì A=x-2y+x-2y=2x-4y
b: \(B=x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
B=x^2+x^2-4=2x^2-4
TH2: -2<=x<=2
B=x^2+4-x^2=4
c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)
d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)
1) ĐKXĐ: \(x>0;x\ne4;x\ne9\)
(*lười lắm, ko chép lại đề nha :V*)
\(P=\frac{\left(2+\sqrt{x}\right)^2+\sqrt{x}\left(2-\sqrt{x}\right)+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\frac{2\sqrt{x}-\left(\sqrt{x}+3\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\\ =\frac{4+4\sqrt{x}+x+2\sqrt{x}-x+4x+2\sqrt{x}-4}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}-3}\\ =\frac{4x+8\sqrt{x}}{2+\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\\ =\frac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{4x}{\sqrt{x}-3}\)
2) Để P>0 thì
\(\frac{4x}{\sqrt{x}-3}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}4x>0\\\sqrt{x}-3>0\end{matrix}\right.\\\left\{{}\begin{matrix}4x< 0\\\sqrt{x}-3< 0\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\\sqrt{x}>3\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\\sqrt{x}< 3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>0\\x>9\end{matrix}\right.\\\left\{{}\begin{matrix}x< 0\\x< 9\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>9\\x< 0\left(ktm\right)\end{matrix}\right.\)
Vậy với \(x>9\) thì \(P>0\).
Chúc bạn học tốt nha
.
Bạn giải thêm cho mk câu này đi
c) tìm giá trị của x để P = -1
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
check giùm mik
a: Thay x=36 vào B, ta được:
\(B=\dfrac{36+2}{36+6+1}=\dfrac{38}{43}\)
b: Ta có: \(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{x-\sqrt{x}+3}{x\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
\(M=\dfrac{\sqrt{x}+5}{\sqrt{x}-2}=\dfrac{\sqrt{x}-2+7}{\sqrt{x}-2}=1+\dfrac{7}{\sqrt{x}-2}\)
Để M nguyên \(\Leftrightarrow\text{ }7\text{ }⋮\text{ }\left(\sqrt{x}-2\right)\)
=> \(\sqrt{x}-2\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{1;3;9\right\}\)
\(\Rightarrow x\in\left\{1;9;81\right\}\)
\(=\sqrt{\left(x+2y^2\right)^2}-\sqrt{\left(2x-3y^2\right)^2}\)\(=x+2y^2-3y^2+2x=3x-y^2=3\cdot\sqrt{2}-1\)