\(\text{Tính }C=\frac{\left(1+\sqrt{3}\right)x^2y-\left(2-\sqrt{5}\right)xy^2}{x^3+y^3}\text{ với }x,y\ne0\text{ và }\frac{x}{4}=\frac{y}{7}\)
Những câu hỏi liên quan
\(\text{cho }xy\ne0\text{ và x + y = 1 }\)
\(\text{Chứng minh rằng}:\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
(chứng minh rằng\) x y 3 −1 - Online Math
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Ta có \(y^3-1=\left(y-1\right)\left(y^2+y+1\right)=-x\left(y^2+y+1\right)\)
(vì \(xy\ne0\Rightarrow x,y\ne0\))
\(\Rightarrow x-1\ne0;y-1\ne0\)
\(\Rightarrow\frac{x}{y^3-1}=\frac{-1}{y^2+y+1}\)
\(x^3-1=\left(x-1\right)\left(x^2-x+1\right)=-y\left(x^2-x+1\right)\Rightarrow\frac{y}{x^3-1}=\frac{-1}{x^2+x+1}\)
\(\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}=\frac{-1}{y^2+y+1}+\frac{-1}{x^2+x+1}\)
\(=-\left(\frac{x^2+x+1+y^2+y+1}{\left(x^2+x+1\right)\left(y^2+y+1\right)}\right)=-\left(\frac{\left(x+y\right)^2-2xy+\left(x+y\right)+2}{x^2y^2+\left(x+y\right)^2-2xy+xy\left(x+y\right)+xy+\left(x+y\right)+1}\right)\)
\(=-\frac{4-2xy}{x^2y^2+3}\Rightarrow\frac{x}{y^3-1}+\frac{y}{x^3-1}-\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
Bài 1: Tính
Asqrt{5-2text{√}6}+sqrt{5+2text{√}6}
B left(sqrt{10}+sqrt{6}right)sqrt{8-2text{√}15}
Csqrt{4+text{√}7}+sqrt{4-text{√}7}
Dleft(3+text{√}5right)left(text{√}10-text{√}2right)sqrt{3-text{√}5}
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a0
b, x-2sqrt{xy}+y left(xge0;yge0right)
c, sqrt{xy}+2text{√}x-3text{√}y-6left(xge0;yge0right)
Bài 3: Rút gọn
M left(frac{1}{text{√}x-1}-frac{1}{text{√}x}right)divleft(frac{text{√}x+1}{text{√}x-2}-frac{text{√}x+2}{text{√}x-1}right)
a, Rút gọn...
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Bài 1: Tính
A=\(\sqrt{5-2\text{√}6}+\sqrt{5+2\text{√}6}\)
B= \(\left(\sqrt{10}+\sqrt{6}\right)\sqrt{8-2\text{√}15}\)
C=\(\sqrt{4+\text{√}7}+\sqrt{4-\text{√}7}\)
D=\(\left(3+\text{√}5\right)\left(\text{√}10-\text{√}2\right)\sqrt{3-\text{√}5}\)
Bài 2: Phân tích thành nhân tử
a, ab+ba+√a+1; a>=0
b, x-2\(\sqrt{xy}\)+y \(\left(x\ge0;y\ge0\right)\)
c, \(\sqrt{xy}+2\text{√}x-3\text{√}y-6\)\(\left(x\ge0;y\ge0\right)\)
Bài 3: Rút gọn
M= \(\left(\frac{1}{\text{√}x-1}-\frac{1}{\text{√}x}\right)\div\left(\frac{\text{√}x+1}{\text{√}x-2}-\frac{\text{√}x+2}{\text{√}x-1}\right)\)
a, Rút gọn M
b, Tính giá trị của M khi x=2
c, Tìm x để M>0
Bài 1:
\(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}=\sqrt{2+3-2\sqrt{2.3}}+\sqrt{2+3+2\sqrt{2.3}}\)
\(=\sqrt{(\sqrt{2}-\sqrt{3})^2}+\sqrt{\sqrt{2}+\sqrt{3})^2}\)
\(=|\sqrt{2}-\sqrt{3}|+|\sqrt{2}+\sqrt{3}|=\sqrt{3}-\sqrt{2}+\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
\(B=(\sqrt{10}+\sqrt{6})\sqrt{8-2\sqrt{15}}\)
\(=(\sqrt{10}+\sqrt{6}).\sqrt{3+5-2\sqrt{3.5}}\)
\(=(\sqrt{10}+\sqrt{6})\sqrt{(\sqrt{5}-\sqrt{3})^2}\)
\(=\sqrt{2}(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\sqrt{2}(5-3)=2\sqrt{2}\)
\(C=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
\(C^2=8+2\sqrt{(4+\sqrt{7})(4-\sqrt{7})}=8+2\sqrt{4^2-7}=8+2.3=14\)
\(\Rightarrow C=\sqrt{14}\)
\(D=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{2}\sqrt{3-\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{6-2\sqrt{5}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{5+1-2\sqrt{5.1}}\)
\(=(3+\sqrt{5})(\sqrt{5}-1).\sqrt{(\sqrt{5}-1)^2}\)
\(=(3+\sqrt{5})(\sqrt{5}-1)^2=(3+\sqrt{5})(6-2\sqrt{5})=2(3+\sqrt{5})(3-\sqrt{5})=2(3^2-5)=8\)
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Bài 2:
a) Bạn xem lại đề.
b) \(x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}.\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2\)
c)
\(\sqrt{xy}+2\sqrt{x}-3\sqrt{y}-6=(\sqrt{x}.\sqrt{y}+2\sqrt{x})-(3\sqrt{y}+6)\)
\(=\sqrt{x}(\sqrt{y}+2)-3(\sqrt{y}+2)=(\sqrt{x}-3)(\sqrt{y}+2)\)
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Bài 3:
a) ĐKXĐ:\(x>0; x\neq 1; x\neq 4\)
\(M=\frac{\sqrt{x}-(\sqrt{x}-1)}{(\sqrt{x}-1)\sqrt{x}}:\frac{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{(x-1)-(x-4)}{(\sqrt{x}-2)(\sqrt{x}-1)}=\frac{1}{\sqrt{x}(\sqrt{x}-1)}:\frac{3}{(\sqrt{x}-2)(\sqrt{x}-1)}\)
\(\frac{1}{\sqrt{x}(\sqrt{x}-1)}.\frac{(\sqrt{x}-2)(\sqrt{x}-1)}{3}=\frac{\sqrt{x}-2}{3\sqrt{x}}\)
b)
Khi $x=2$ \(M=\frac{\sqrt{2}-2}{3\sqrt{2}}=\frac{1-\sqrt{2}}{3}\)
c)
Để \(M>0\leftrightarrow \frac{\sqrt{x}-2}{3\sqrt{x}}>0\leftrightarrow \sqrt{x}-2>0\leftrightarrow x>4\)
Kết hợp với ĐKXĐ suy ra $x>4$
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giải hpt:1)begin{cases}text{x+y+xy(2x+y)5xy }text{x+y+xy(3x-y)4xy}end{cases} 2)begin{cases}left(2x+y+1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end{cases} 3)begin{cases}sqrt{9x+frac{y}{x}}+2.sqrt{y+frac{2x}{y}}4left(frac{2x}{y^2}-1right)left(frac{y}{x^2}-9right)18end{cases}
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giải hpt:1)\(\begin{cases}\text{x+y+xy(2x+y)=5xy }\\\text{x+y+xy(3x-y)=4xy}\end{cases}\)
2)\(\begin{cases}\left(2x+y+1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{cases}\)
3)\(\begin{cases}\sqrt{9x+\frac{y}{x}}+2.\sqrt{y+\frac{2x}{y}}=4\\\left(\frac{2x}{y^2}-1\right)\left(\frac{y}{x^2}-9\right)=18\end{cases}\)
1. \(\begin{cases}x+y+xy\left(2x+y\right)=5xy\\x+y+xy\left(3x-y\right)=4xy\end{cases}\) \(\Leftrightarrow\begin{cases}2y-x=1\\x+y+xy\left(2x+y\right)=5xy\end{cases}\) (trừ 2 vế cho nhau)
\(\Leftrightarrow\begin{cases}x=2y-1\\\left(2y-1\right)+y+\left(2y-1\right)y\left(4y-2+y\right)=5\left(2y-1\right)y\end{cases}\) \(\Leftrightarrow\begin{cases}x=2y-1\\10y^3-19y^2+10y-1=0\end{cases}\) \(\Leftrightarrow\begin{cases}x=1\\y=1\end{cases}\)
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Giải hệ phương trình:
a,left{{}begin{matrix}text{2
x
−
7
y
20
}text{3
x
+
7
y
−
5}end{matrix}right.
b,left{{}begin{matrix}text{2
x
+
3
y
8
}text{−
3
x
+
5
y
7
}end{matrix}right.
c,left{{}begin{matrix}-2x+frac{1}{2}y3text{5
x
+
3
y
11
}end{matrix}right.
d,left{{}begin{matrix}text{(
x
−
16
)(
y
+
6
)
x
y
−
36
}text{(
x
+
8
)(
y
−
3
)
x
y
−
54
}end{matrix}right.
e,left{{}begin{matrix}text{3
x
−
|
y
|
1
}text{5
x
+
3
y
11
}end{matrix}right.
f,left{{}begin{matrix}frac{3}{x}+frac{4}{y}...
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Giải hệ phương trình:
a,\(\left\{{}\begin{matrix}\text{2 x − 7 y = 20 }\\\text{3 x + 7 y = − 5}\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}\text{2 x + 3 y = 8 }\\\text{− 3 x + 5 y = 7 }\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}-2x+\frac{1}{2}y=3\\\text{5 x + 3 y = 11 }\end{matrix}\right.\)
d,\(\left\{{}\begin{matrix}\text{( x − 16 )( y + 6 ) = x y − 36 }\\\text{( x + 8 )( y − 3 ) = x y − 54 }\end{matrix}\right.\)
e,\(\left\{{}\begin{matrix}\text{3 x − | y | = 1 }\\\text{5 x + 3 y = 11 }\end{matrix}\right.\)
f,\(\left\{{}\begin{matrix}\frac{3}{x}+\frac{4}{y}=5\\\frac{1}{x}-\frac{1}{y}=1\end{matrix}\right.\)
g,\(\left\{{}\begin{matrix}x+2\sqrt{y-1}=3\\3x+4\sqrt{y-1}=7\end{matrix}\right.\)
Giúp mình với ạ!!!
Rút gọn các biểu thức sau:
a) $A=4 \sqrt{x^{2}+1}-2 \sqrt{16\left(x^{2}+1\right)}+5 \sqrt{25\left(x^{2}+1\right)} \text {; }$
b) $B=\dfrac{2}{x+y} \sqrt{\dfrac{3(x+y)^{2}}{4}}$ với $x+y>0$;
c) $C=\dfrac{3}{3 a-1} \sqrt{5 a\left(1-6 a+a^{2}\right)}$ với $a>\frac{1}{3}$.
a) \(A=4\sqrt{x^2+1}-2\sqrt{16\left(x^2+1\right)}+5\sqrt{25\left(x^2+1\right).}\)
\(=4\sqrt{x^2+1}-2.4\sqrt{x^2+1}+5.5\sqrt{x^2+1}\)
\(=4\sqrt{x^2+1}-8\sqrt{x^2+1}+25\sqrt{x^2+1}\)
\(=\left(4-8+25\right)\sqrt{x^2+1}\)
\(=21\sqrt{x^2+1}\)
b) \(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}\)
\(B=\frac{2}{x+y}.\frac{\sqrt{3}\left(x+y\right)}{2}\)
\(B=\frac{\sqrt{3}\left(x+y\right)}{x+y}\)
\(B=\sqrt{3}\)
Dạ đậy ạ,mong dc gp
Xem thêm câu trả lời
cho x,y,z là các số dương thỏa \(x^2+y^2+z^2=3\)
chứng minh:\(\frac{x^2}{y+2\text{z}}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}+\frac{1}{1+\sqrt{3+2\left(xy+yz+x\text{z}\right)}}\ge\frac{5}{4}\)
+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)
\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)
+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)
Áp dụng bđt Bunhiacopxki
\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)
\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)
(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Tương tự: \(xy^2+yz^2+zx^2\le3\)
\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)
\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)
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Ai giải giúp mấy bài toán vsBài 1:Asqrt{frac{1}{text{√}2+1}-frac{text{√}8-text{√}10}{2-text{√}5}}Bfrac{5text{√}5}{text{√}5+2}+frac{text{√}5}{text{√}5-1}-frac{3text{√}5}{3+text{√}5}Bài 2 rút gọn biểu thứcAleft(frac{x+sqrt[]{xy}}{text{√}x+text{√}y}-2right):frac{1}{text{√}x+2} với x :y 0Bleft(frac{a}{a-2text{√}a}+frac{a}{text{√}a-2}right):frac{text{√}a+1}{a-4text{√}a+4}Bài 3 cho biểu thứcPleft(frac{x-2}{x+2text{√}x}+frac{1}{text{√}x+2}right)frac{text{√}x+1}{text{√}x-1}a)Rút gọn Pb)tìm x để Ptext{√}...
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Ai giải giúp mấy bài toán vs
Bài 1:
A=\(\sqrt{\frac{1}{\text{√}2+1}-\frac{\text{√}8-\text{√}10}{2-\text{√}5}}\)
B=\(\frac{5\text{√}5}{\text{√}5+2}+\frac{\text{√}5}{\text{√}5-1}-\frac{3\text{√}5}{3+\text{√}5}\)
Bài 2 rút gọn biểu thức
A=\(\left(\frac{x+\sqrt[]{xy}}{\text{√}x+\text{√}y}-2\right):\frac{1}{\text{√}x+2}\) với x :y >0
B=\(\left(\frac{a}{a-2\text{√}a}+\frac{a}{\text{√}a-2}\right):\frac{\text{√}a+1}{a-4\text{√}a+4}\)
Bài 3 cho biểu thức
P=\(\left(\frac{x-2}{x+2\text{√}x}+\frac{1}{\text{√}x+2}\right)\frac{\text{√}x+1}{\text{√}x-1}\)
a)Rút gọn P
b)tìm x để P=\(\text{√}x+\frac{5}{2}\)
bài 4 rút gọn biểu thức
A=\(\frac{1}{x+\text{√}x}+\frac{2\text{√}x}{x-1}-\frac{1}{x-\text{√}x}\)
B=\(\left(\frac{x}{x+3\text{√}x}+\frac{1}{\text{√}x+3}\right):\left(1-\frac{2}{\text{√}x}+\frac{6}{x+3\text{√}x}\right)\)
Bài 5
A=\(\left(\frac{2}{\text{√}x-3}-\frac{1}{\text{√}x+3}-\frac{x}{\text{√}x\left(x-9\right)}\right):\text{(√}x+3-\frac{x}{\text{√}x-3}\)
a)rút gọn A
b)tìm gtri x để A= -1/4
AI GIẢI GIÙM MÌNH ĐI MÌNH TẠ ƠN
Tìm GTLN của:
\(A=\frac{1}{x^3}+\frac{1}{y^3}\text{ biết }x,y\ne0\text{ và }\left(x+y+1\right)xy=x^2+y^2\)
đặt Afrac{sqrt{yz}}{x+3sqrt{yz}}+frac{sqrt{zx}}{y+3sqrt{zx}}+frac{sqrt{xy}}{z+3sqrt{xy}}Rightarrow1-3Afrac{x}{x+3sqrt{yz}}+frac{y}{y+3sqrt{zx}}+frac{z}{z+3sqrt{xy}}gefrac{x}{x+frac{3}{2}left(y+zright)}+frac{y}{y+frac{3}{2}left(z+xright)}+frac{z}{z+frac{3}{2}left(x+yright)}frac{2x}{2x+3left(y+zright)}+frac{2y}{2y+3left(z+xright)}+frac{2z}{2z+3left(x+yright)}frac{2x^2}{2x^2+3xy+3xz}+frac{2y^2}{2y^2+3yz+3xy}+frac{2z^2}{2z^2+3zx+3yz}gefrac{2left(x+y+zright)^2}{2left(x^2+y^2+z^2right)+6left(xy+yz+zxr...
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đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
