Tìm a,b,c biết :
\(\left\{{}\begin{matrix}a+b+c=270\\a=2.b\\b=\frac{1}{3}.c\end{matrix}\right.\)
Những câu hỏi liên quan
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x+y+z=9\\\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\\zy+yz+zx=27\end{matrix}\right.\)
a.
\(\frac{3x-36}{12}=\frac{5y-45}{15}=\frac{z-1}{1}=\frac{3x+5y-z-50}{26}=\frac{-48}{26}\)
\(\Rightarrow\frac{x-12}{4}=\frac{-48}{26}\Rightarrow x=...\)
Tương tự với y, z, nhưng chắc bạn nhầm đề, nếu pt bên dưới là -2 thì nó ra \(\frac{-52}{26}=-2\) kết quả đẹp hơn nhiều
b. Không rõ đề
c.
\(x+y+z=9\Rightarrow\left(x+y+z\right)^2=81=3.27=3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)
\(\Leftrightarrow x=y=z\Rightarrow\frac{3}{x}=1\Rightarrow x=y=z=3\)
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
Lời giải:
a)
Đặt \(\frac{x-12}{4}=\frac{y-9}{3}=z-1=k\Rightarrow \left\{\begin{matrix} x=4k+12\\ y=3k+9\\ z=k+1\end{matrix}\right.\)
Khi đó:
\(3x+5y-z=2\)
\(\Leftrightarrow 3(4k+12)+5(3k+9)-(k+1)=2\)
$\Rightarrow k=-3$
\(\Rightarrow \left\{\begin{matrix} x=4k+12=0\\ y=3k+9=0\\ z=k+1=-2\end{matrix}\right.\)
b)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}=\frac{a+b+b+c+c+a}{6+7+8}=\frac{2(a+b+c)}{21}=\frac{2.14}{21}=\frac{4}{3}\)
\(\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\\ a+b+c=14\end{matrix}\right.\Rightarrow \left\{\begin{matrix} c=6\\ a=\frac{14}{3}\\ b=\frac{10}{3}\end{matrix}\right.\)
Lời giải:
a)
Đặt \(\frac{x-12}{4}=\frac{y-9}{3}=z-1=k\Rightarrow \left\{\begin{matrix} x=4k+12\\ y=3k+9\\ z=k+1\end{matrix}\right.\)
Khi đó:
\(3x+5y-z=2\)
\(\Leftrightarrow 3(4k+12)+5(3k+9)-(k+1)=2\)
$\Rightarrow k=-3$
\(\Rightarrow \left\{\begin{matrix} x=4k+12=0\\ y=3k+9=0\\ z=k+1=-2\end{matrix}\right.\)
b)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}=\frac{a+b+b+c+c+a}{6+7+8}=\frac{2(a+b+c)}{21}=\frac{2.14}{21}=\frac{4}{3}\)
\(\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} a+b=8\\ b+c=\frac{28}{3}\\ c+a=\frac{32}{3}\\ a+b+c=14\end{matrix}\right.\Rightarrow \left\{\begin{matrix} c=6\\ a=\frac{14}{3}\\ b=\frac{10}{3}\end{matrix}\right.\)
11 tháng 2 2020 lúc 21:04
1. left{{}begin{matrix}a,b,c0a+b+c1end{matrix}right.. Cmr: frac{ab}{sqrt{left(1-cright)^2left(1+cright)}}+frac{bc}{sqrt{left(1-aright)^2left(1+aright)}}+frac{ca}{sqrt{left(1-bright)^3left(1+bright)}}lefrac{3sqrt{2}}{8}
2. left{{}begin{matrix}a,b,c0a+b+cle1end{matrix}right.. Cmr: frac{1}{a^2+b^2+c^2}+frac{1}{ableft(a+bright)}+frac{1}{bcleft(b+cright)}+frac{1}{acleft(a+cright)}gefrac{87}{2}
3. left{{}begin{matrix}a,b,c0ab+bc+ca2abcend{matrix}right.. Cmr: frac{1}{aleft(2a-1right)^2}+frac{1}{bleft...
Đọc tiếp
1. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}+\frac{bc}{\sqrt{\left(1-a\right)^2\left(1+a\right)}}+\frac{ca}{\sqrt{\left(1-b\right)^3\left(1+b\right)}}\le\frac{3\sqrt{2}}{8}\)
2. \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c\le1\end{matrix}\right.\). Cmr: \(\frac{1}{a^2+b^2+c^2}+\frac{1}{ab\left(a+b\right)}+\frac{1}{bc\left(b+c\right)}+\frac{1}{ac\left(a+c\right)}\ge\frac{87}{2}\)
3. \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca=2abc\end{matrix}\right.\). Cmr: \(\frac{1}{a\left(2a-1\right)^2}+\frac{1}{b\left(2b-1\right)^2}+\frac{1}{c\left(2c-1\right)^2}\ge\frac{1}{2}\)
4. \(\left\{{}\begin{matrix}x,y,z>0\\x+y+z=2015\end{matrix}\right.\). Tìm min \(A=\frac{x^4+y^4}{x^3+y^3}+\frac{y^4+z^4}{y^3+z^3}+\frac{z^4+x^4}{z^2+x^2}\)
Mn giúp mk với ạ! Thanks nhiều
Mới nghĩ ra 3 câu:
a/ \(\frac{ab}{\sqrt{\left(1-c\right)^2\left(1+c\right)}}=\frac{ab}{\sqrt{\left(a+b\right)^2\left(1+c\right)}}\le\frac{ab}{2\sqrt{ab\left(1+c\right)}}=\frac{1}{2}\sqrt{\frac{ab}{1+c}}\)
\(\sum\sqrt{\frac{ab}{1+c}}\le\sqrt{2\sum\frac{ab}{1+c}}\)
\(\sum\frac{ab}{1+c}=\sum\frac{ab}{a+c+b+c}\le\frac{1}{4}\sum\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)=\frac{1}{4}\)
c/ \(ab+bc+ca=2abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\Rightarrow x+y+z=2\)
\(VT=\sum\frac{x^3}{\left(2-x\right)^2}\)
Ta có đánh giá: \(\frac{x^3}{\left(2-x\right)^2}\ge x-\frac{1}{2}\) \(\forall x\in\left(0;2\right)\)
\(\Leftrightarrow2x^3\ge\left(2x-1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow9x^2-12x+4\ge0\Leftrightarrow\left(3x-2\right)^2\ge0\)
d/ Ta có đánh giá: \(\frac{x^4+y^4}{x^3+y^3}\ge\frac{x+y}{2}\)
\(\Leftrightarrow\left(x-y\right)^2\left(x^2+xy+y^2\right)\ge0\)
Akai Haruma, Nguyễn Ngọc Lộc , @tth_new, @Băng Băng 2k6, @Trần Thanh Phương, @Nguyễn Việt Lâm
Mn giúp e vs ạ! Thanks!
\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
GIẢI CÁC HPT SAU:
a) left{{}begin{matrix}3x-5ge2x+12frac{x-3}{2}lefrac{2x+27}{3}end{matrix}right.
b) left{{}begin{matrix}2-7xge x-143x+1ge6x-11end{matrix}right.
c) left{{}begin{matrix}3x+frac{3}{5} x+22x+3
frac{7x-3}{4}end{matrix}right.
d) left{{}begin{matrix}frac{4x+5}{2} 4x-3xleft(x-1right)geleft(x-3right)left(4+xright)end{matrix}right.
GIÚP MIK VỚI
Đọc tiếp
GIẢI CÁC HPT SAU:
a) \(\left\{{}\begin{matrix}3x-5\ge2x+12\\\frac{x-3}{2}\le\frac{2x+27}{3}\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}2-7x\ge x-14\\3x+1\ge6x-11\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}3x+\frac{3}{5}< x+2\\2x+3 >\frac{7x-3}{4}\end{matrix}\right.\)
d) \(\left\{{}\begin{matrix}\frac{4x+5}{2}< 4x-3\\x\left(x-1\right)\ge\left(x-3\right)\left(4+x\right)\end{matrix}\right.\)
GIÚP MIK VỚI
Giải hệ pt sau
a)\(\left\{{}\begin{matrix}\frac{x-12}{4}=\frac{y-9}{3}=z-1\\3x+5y-z=2\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\frac{a+b}{6}=\frac{b+c}{7}=\frac{a+c}{8}\\a+b+c=14\end{matrix}\right.\)
2 tháng 3 2020 lúc 23:45
1. a) left{{}begin{matrix}x,y,z0xyz1end{matrix}right.. Tìm max Pfrac{1}{sqrt{x^5-x^2+3xy+6}}+frac{1}{sqrt{y^5-y^2+3yz+6}}+frac{1}{sqrt{z^5-z^2+zx+6}}
b) left{{}begin{matrix}x,y,z0xyz8end{matrix}right.. Min Pfrac{x^2}{sqrt{left(1+x^3right)left(1+y^3right)}}+frac{y^2}{sqrt{left(1+y^3right)left(1+z^3right)}}+frac{z^2}{sqrt{left(1+z^3right)left(1+x^3right)}}
c) x,y,z0. Min Psqrt{frac{x^3}{x^3+left(y+zright)^3}}+sqrt{frac{y^3}{y^3+left(z+xright)^3}}+sqrt{frac{z^3}{z^3+left(x+yright)^3}}
d) a,b,c0;...
Đọc tiếp
1. a) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Tìm max \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+zx+6}}\)
b) \(\left\{{}\begin{matrix}x,y,z>0\\xyz=8\end{matrix}\right.\). Min \(P=\frac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\frac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\frac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
c) \(x,y,z>0.\) Min \(P=\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}+\sqrt{\frac{y^3}{y^3+\left(z+x\right)^3}}+\sqrt{\frac{z^3}{z^3+\left(x+y\right)^3}}\)
d) \(a,b,c>0;a^2+b^2+c^2+abc=4.Cmr:2a+b+c\le\frac{9}{2}\)
e) \(\left\{{}\begin{matrix}a,b,c>0\\a+b+c=3\end{matrix}\right.\). Cmr: \(\frac{a}{b^3+ab}+\frac{b}{c^3+bc}+\frac{c}{a^3+ca}\ge\frac{3}{2}\)
f) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=4\end{matrix}\right.\) Cmr: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le3\)
g) \(\left\{{}\begin{matrix}a,b,c>0\\ab+bc+ca+abc=2\end{matrix}\right.\) Max : \(Q=\frac{a+1}{a^2+2a+2}+\frac{b+1}{b^2+2b+2}+\frac{c+1}{c^2+2c+2}\)
Câu c quen thuộc, chém trước:
Ta có BĐT phụ: \(\frac{x^3}{x^3+\left(y+z\right)^3}\ge\frac{x^4}{\left(x^2+y^2+z^2\right)^2}\) \((\ast)\)
Hay là: \(\frac{1}{x^3+\left(y+z\right)^3}\ge\frac{x}{\left(x^2+y^2+z^2\right)^2}\)
Có: \(8(y^2+z^2) \Big[(x^2 +y^2 +z^2)^2 -x\left\{x^3 +(y+z)^3 \right\}\Big]\)
\(= \left( 4\,x{y}^{2}+4\,x{z}^{2}-{y}^{3}-3\,{y}^{2}z-3\,y{z}^{2}-{z}^{3 } \right) ^{2}+ \left( 7\,{y}^{4}+8\,{y}^{3}z+18\,{y}^{2}{z}^{2}+8\,{z }^{3}y+7\,{z}^{4} \right) \left( y-z \right) ^{2} \)
Từ đó BĐT \((\ast)\) là đúng. Do đó: \(\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\frac{x^2}{x^2+y^2+z^2}\)
\(\therefore VT=\sum\sqrt{\frac{x^3}{x^3+\left(y+z\right)^3}}\ge\sum\frac{x^2}{x^2+y^2+z^2}=1\)
Done.
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Câu 1 chuyên phan bội châu
câu c hà nội
câu g khoa học tự nhiên
câu b am-gm dựa vào hằng đẳng thử rồi đặt ẩn phụ
câu f đặt \(a=\frac{2m}{n+p};b=\frac{2n}{p+m};c=\frac{2p}{m+n}\)
Gà như mình mấy câu còn lại ko bt nha ! để bạn tth_pro full cho nhé !
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Nguyễn Ngọc Lộc , ?Amanda?, Phạm Lan Hương, Akai Haruma, @Trần Thanh Phương, @Nguyễn Việt Lâm,
Giúp em vs ạ! Thanks nhiều ạ
1.) liệt kê các tập hợp sau :
a.) A left{{}begin{matrix}end{matrix}right.xin N|}2le xle10left{right}
b.) B left{{}begin{matrix}end{matrix}right.xin Z|9le x^2le36left{right}}
c.) C left{{}begin{matrix}end{matrix}right.nin N}^{cdot}|3le n^2le30left{right}
B.) B là tập hợp các số thực x thỏa x2 - 4x +2 0
d.) D left{{}begin{matrix}end{matrix}right.frac{1}{n+1}}|nin N;nle4left{right}
e.) E left{{}begin{matrix}end{matrix}right.2n^2-1|nin N^{cdot}},nle7left{right}
2.) chỉ ra tính chất đặc t...
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1.) liệt kê các tập hợp sau :
a.) A = \(\left\{{}\begin{matrix}\\\end{matrix}\right.x\in N|}2\le x\le10\left\{\right\}\)
b.) B =\(\left\{{}\begin{matrix}\\\end{matrix}\right.x\in Z|9\le x^2\le36\left\{\right\}}\)
c.) C = \(\left\{{}\begin{matrix}\\\end{matrix}\right.n\in N}^{\cdot}|3\le n^2\le30\left\{\right\}\)
B.) B là tập hợp các số thực x thỏa x2 - 4x +2 = 0
d.) D = \(\left\{{}\begin{matrix}\\\end{matrix}\right.\frac{1}{n+1}}|n\in N;n\le4\left\{\right\}\)
e.) E = \(\left\{{}\begin{matrix}\\\end{matrix}\right.2n^2-1|n\in N^{\cdot}},n\le7\left\{\right\}\)
2.) chỉ ra tính chất đặc trưng :
a.) A = \(\left\{{}\begin{matrix}\\\end{matrix}\right.0;1;2;3;4\left\{\right\}}\)
b.) B = \(\left\{{}\begin{matrix}\\\end{matrix}\right.0;4;8;12;16\left\{\right\}}\)
c.) C = \(\left\{{}\begin{matrix}\\\end{matrix}\right.0;4;9;16;25;36\left\{\right\}}\)
3.) Trong các tập hợp sau , tập hợp nào là con tập nào :
a.) A = \(\left\{{}\begin{matrix}\\\end{matrix}\right.1;2;3\left\{\right\}}\)
B = \(\left\{{}\begin{matrix}\\\end{matrix}\right.x\in N^{\cdot}|n\le4\left\{\right\}}\)
b.) A = \(\left\{{}\begin{matrix}\\\end{matrix}\right.n\in N^{\cdot}}|n\le5\left\{\right\}\)
B = \(\left\{{}\begin{matrix}\\\end{matrix}\right.n\in Z|0\le|n|\le5\left\{\right\}}\)
Tìm a,b,c biết :
\(\left\{{}\begin{matrix}a+b+c=90\\\frac{2}{3}a=\frac{1}{4}b\\\frac{1}{2}b=\frac{1}{3}c\end{matrix}\right.\)
Ta có: \(\frac{2}{3}a=\frac{1}{4}b\)
\(\Leftrightarrow\frac{2a}{3}=\frac{b}{4}\)
\(\Leftrightarrow2a=\frac{3b}{4}\)
hay \(a=\frac{3b}{4}:2=\frac{3b}{8}\)
Ta có: \(\frac{1}{2}b=\frac{1}{3}c\)
\(\Leftrightarrow\frac{b}{2}=\frac{c}{3}\)
hay \(c=\frac{3b}{2}\)
Ta có: a+b+c=90
\(\Leftrightarrow\frac{3b}{8}+b+\frac{3b}{2}=90\)
\(\Leftrightarrow b\left(\frac{3}{8}+1+\frac{3}{2}\right)=90\)
\(\Leftrightarrow b\cdot\frac{23}{8}=90\)
hay \(b=90:\frac{23}{8}=\frac{720}{23}\)
Ta có: \(a=\frac{3b}{8}\)(cmt)
hay \(a=3\cdot\frac{720}{23}:8=\frac{270}{23}\)
Ta có: a+b+c=90
\(\Leftrightarrow c=90-a-b=90-\frac{270}{23}-\frac{720}{23}=\frac{1080}{23}\)
Vậy: \(\left(a,b,c\right)=\left(\frac{270}{23};\frac{720}{23};\frac{1080}{23}\right)\)
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