Cho
\(a=x+\frac{1}{x}\left(x\ne0\right)\)
hay tinh gia tri cua
\(A=x^{13}+\frac{1}{x^{13}}\) theo a
Những câu hỏi liên quan
cau 1: tinh gia tri cua x thoa manleft(x-3right)left(x^2+3x+9right)+xleft(x+2sqrt{2}right)left(2sqrt{2}-xright)-3cau 2.tinh GTLN cua bieu thuc2x-2x^2+13cau 3. tinh gia tri cua bieu thuc frac{3^{left(x+yright)^2}}{3^{left(x-yright)^2}}voi xyfrac{1}{2}cau 4. tim GTLN cua-3x^2-6x-4cau 5. cho ham so : f(x)frac{1}{5x+9}tinh gia tri cua fleft(frac{40}{25}right)cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
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cau 1: tinh gia tri cua x thoa man
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\sqrt{2}\right)\left(2\sqrt{2}-x\right)=-3\)
cau 2.tinh GTLN cua bieu thuc
\(2x-2x^2+13\)
cau 3. tinh gia tri cua bieu thuc
\(\frac{3^{\left(x+y\right)^2}}{3^{\left(x-y\right)^2}}\)voi xy=\(\frac{1}{2}\)
cau 4. tim GTLN cua
\(-3x^2-6x-4\)
cau 5. cho ham so : f(x)=\(\frac{1}{5x+9}\)
tinh gia tri cua \(f\left(\frac{40}{25}\right)\)
cau 6. cho hinh thang can ABCD . Day nho AB,goc D bang 64 do. tinh so do goc ngoai tai A
1) Cho bieu thuc: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\left(x\ge0,x\ne16\right)\)
a) Cho bieu thuc A= \(\frac{\sqrt{x}+4}{\sqrt{x}+2}\) ; voi cac cua bieu thuc A va B da cho, hay tim cac gia tri cua x nguyen de gia tri cua bieu thuc B(A;-1) la so nguyen
1) A frac{frac{3}{4}-frac{3}{11}+frac{3}{13}}{frac{5}{7}-frac{5}{11}+frac{5}{13}}+frac{frac{1}{2}-frac{1}{3}+frac{1}{4}}{frac{5}{4}-frac{5}{6}+frac{5}{8}}b) Cho 3 so x,y,z la 3 so khac 0 thoa man dieu kien :frac{y+z-x}{x}frac{z+x-y}{y}frac{x+y-z}{z}Hay tinh gia tri bieu thuc:Bleft(1+frac{x}{y}right)left(1+frac{y}{z}right)left(1+frac{z}{x}right)
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1) A= \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
b) Cho 3 so x,y,z la 3 so khac 0 thoa man dieu kien :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Hay tinh gia tri bieu thuc:\(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
Bài 1 :
Ta có :
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(A=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(A=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(A=\frac{3}{5}+\frac{2}{5}\)
\(A=1\)
\(b)\) Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Đo đó :
\(\frac{y+z-x}{x}=2\)\(\Rightarrow\)\(y+z=3x\)\(\left(1\right)\)
\(\frac{z+x-y}{y}=2\)\(\Rightarrow\)\(x+z=3y\)\(\left(2\right)\)
\(\frac{x+y-z}{z}=2\)\(\Rightarrow\)\(x+y=3z\)\(\left(3\right)\)
Lại có : \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)
Thay (1), (2) và (3) vào \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\) ta được :
\(B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)
Vậy \(B=8\)
Chúc bạn học tốt ~
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bạn phùng minh quân câu 1 a tại sao lại rút gọn được \(\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\) vậy nó không cùng nhân tử mà
câu b \(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{\left(y-y+y\right)+\left(-x+x+x\right)+\left(z+z-z\right)}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)sao lại ra bằng 2
(mình chỉ góp ý thôi nha tại mình làm thấy nó sai sai)
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Cho bieu thuc \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2016}{x}\)
a, Voi gia tri nguyen nao cua x thi bieu thuc A co gia tri nguyen
b,Voi gia tri nao cua x thi A co gia tri duong
cho bieu thuc A = \(\left(\frac{x-\sqrt{x}}{\sqrt{x}-1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
a. tim x de bieu thuc A co nghia ?rut gon A ?
b. tinh gia tri cua bieu thuc A tai x=7+4√3
a. A có nghĩa khi \(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne\\\frac{x+\sqrt{x}}{\sqrt{x}+1}\ne0\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
A\(=\frac{x-\sqrt{x}+\sqrt{x}-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{x+\sqrt{x}}\)\(=\frac{x-1}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}.\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}}\)
b. \(x=7+4\sqrt{3}\Rightarrow\)A = \(\frac{\sqrt{7+4\sqrt{3}}+1}{\sqrt{7+4\sqrt{3}}}=\frac{\sqrt{\left(2+\sqrt{3}\right)^2}+1}{\sqrt{\left(2+\sqrt{3}\right)^2}}=\frac{3+\sqrt{3}}{2+\sqrt{3}}\)
cho bieu thuc A=\(\left(\frac{x-\sqrt{x}}{\sqrt{x}+1}+1\right):\left(\frac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
(x≥0;x≠1)
a.Tim x de bieu thuc A co nghia ? rut gon A ?
b. Tinh gia tri cua bieu thuc A tai x=7=4√3
cho so huu ti \(x=\frac{a+17}{a}\left(a\ne0\right)\) , voi gia tri nguyen nao cua a thi x la so nguyen
\(x=\frac{a+17}{a}=\frac{a}{a}+\frac{17}{a}=1+\frac{17}{a}.\)
Để x là số nguyên thì \(\frac{17}{a}\)phải là số nguyên
\(\Rightarrow a\inƯ\left(17\right)=\left\{1;17;-1;-17\right\}\)
Vậy nếu \(\Rightarrow a\in\left\{1;17;-1;-17\right\}\)thì x là số nguyên
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cho a + b = 13 hay tinh gia tri cua bieu thuc 102 x a + 98 x b + 52 x a + 57 x b + a
102xa+98xb+52xa+57xb+a
\(=a\times\left(102+52+1\right)+b\times\left(98+57\right)\)
=155xa+155xb
=155\(\times\)(a+b)
=155x13
=2015
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Cho x,y,z khác 0 và x-y-z=0 . tinh gia tri cua bieu thuc \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
