3a+4b-3c=4. Tìm GTNN của biểu thức : A = 2a+3b-4c? ... Cho a;b;c là các số không âm thỏa mãn:2a+b=6-3c;3a+4b=3c+4.Tìm min ... T = a −2 b 2 a − b +2 a −3 b 2 a + b. Đọc tiếp. ..... cho a và b là hai số thực thỏa mãn 4a + b = 5ab và 2a>b>0.

\(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)\(\Rightarrow a+3b-5c=-2\)

\(\Rightarrow3b=-2+5c-a\)\(\Rightarrow3b+2a-4c=-2+5c-a+2a-4c\)

\(\Rightarrow P=-2+a+c\)

Lại có : \(2a+b+2c=6\Rightarrow2\left(a+c\right)\le6\)

\(\Rightarrow a+c\le3\)

\(\Rightarrow P\le-2+3=1\Rightarrow P\le1\)

Dấu " = " sảy ra \(\Leftrightarrow\hept{\begin{cases}b=0\\3a-3c=4\\2a+2c=6\end{cases}}\)\(\Rightarrow\hept{\begin{cases}b=0\\3a-3c=4\\3a+3c=9\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}a=\frac{13}{6}\\b=0\\c=\frac{5}{6}\end{cases}}\)

Chị chỉ tìm được Max thui 

\(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)

<=> \(\hept{\begin{cases}b+2c=6-2a\\4b-3c=4-3a\end{cases}}\)

<=> \(\hept{\begin{cases}c=\frac{20}{11}-\frac{5a}{11}\\b=\frac{26}{11}-\frac{12}{11}a\end{cases}}\)

P = \(2a+3\left(\frac{26}{11}-\frac{12}{11}a\right)-4\left(\frac{20}{11}-\frac{5a}{11}\right)\)

\(=-\frac{2}{11}+\frac{6}{11}a\ge-\frac{2}{11}\)

Dấu "=" xảy ra <=> a = 0 => c =20/11 và b = 26/11

Vậy min P = -2/11 tại a = 0; b = 26/11 và c= 20/11

Cách tìm max khác:

Ta có: \(\hept{\begin{cases}2a+b+2c=6\\3a+4b-3c=4\end{cases}}\)

<=> \(\hept{\begin{cases}2a+2c=6-b\\3a-3c=4-4b\end{cases}}\) <=> \(\hept{\begin{cases}a+c=3-\frac{b}{2}\\a-c=\frac{4}{3}-\frac{4b}{3}\end{cases}}\)

<=> \(\hept{\begin{cases}a=\frac{13}{6}-\frac{11b}{12}\\c=\frac{5}{6}+\frac{5}{12}b\end{cases}}\)

khi đó P = \(2\left(\frac{13}{6}-\frac{11b}{12}\right)+3b-4\left(\frac{5}{6}+\frac{5}{12}b\right)=1-\frac{1}{2}b\le1\)

Dấu bằng xảy ra khi và chỉ khi b = 0 khi đó a = 13/6 và c = 5/6( thỏa mãn)

Vậy maxP = 1 tại a = 13/6 ;  b = 0 ; c = 5/6.

Sẵn tiện mk chỉ cho bn luôn dạng này nhé.

Phân tích:

Với \(\alpha,\beta,\gamma>0\) thỏa \(\alpha< 2,\beta< 3,\gamma< 4\) ta có:

\(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=\left[\left(2-\alpha\right)a+\dfrac{3}{a}\right]+\left[\left(3-\beta\right)b+\dfrac{9}{2b}\right]+\left[\left(4-\gamma\right)c+\dfrac{4}{c}\right]+\left(\alpha a+\beta b+\gamma c\right)\)

\(\ge2\sqrt{3.\left(2-\alpha\right)}+2\sqrt{\dfrac{9}{2}.\left(3-\beta\right)}+2\sqrt{4.\left(4-\gamma\right)}+\left(\alpha a+\beta b+\gamma c\right)\)

Chọn \(\alpha,\beta,\gamma\) (thỏa đk trên) sao cho:

\(\left\{{}\begin{matrix}\left(2-\alpha\right)a=\dfrac{3}{a}\\\left(3-\beta\right)b=\dfrac{9}{2b}\\\left(4-\gamma\right)c=\dfrac{4}{c}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{2\left(3-\beta\right)}}\\c=\sqrt{\dfrac{4}{\left(4-\gamma\right)}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=\sqrt{\dfrac{3}{2-\alpha}}\\b=\sqrt{\dfrac{9}{6-4\alpha}}\\c=\sqrt{\dfrac{4}{4-3\alpha}}\\\alpha=\dfrac{\beta}{2}=\dfrac{\gamma}{3}\end{matrix}\right.\)

Ta có: \(a+2b+3c\ge20\). Xác định điểm rơi: \(a+2b+3c=20\)

\(\Rightarrow\sqrt{\dfrac{3}{2-\alpha}}+2\sqrt{\dfrac{9}{6-4\alpha}}+3\sqrt{\dfrac{4}{4-3\alpha}}=20\)

Giải ra ta có \(\alpha=\dfrac{5}{4}\Rightarrow\beta=\dfrac{5}{2};\gamma=\dfrac{15}{4}\)

Lời giải:

Ta có: \(A=2a+3b+4c+\dfrac{3}{a}+\dfrac{9}{2b}+\dfrac{4}{c}\)

\(=\left(\dfrac{3a}{4}+\dfrac{3}{a}\right)+\left(\dfrac{b}{2}+\dfrac{9}{2b}\right)+\left(\dfrac{c}{4}+\dfrac{4}{c}\right)+\left(\dfrac{5a}{4}+\dfrac{5b}{2}+\dfrac{15c}{4}\right)\)

\(\ge^{Cauchy}2\sqrt{\dfrac{3a}{4}.\dfrac{3}{a}}+2\sqrt{\dfrac{b}{2}.\dfrac{9}{2b}}+2\sqrt{\dfrac{c}{4}.\dfrac{4}{c}}+\dfrac{5}{4}\left(a+2b+3c\right)\)

\(=3+3+2+\dfrac{5}{4}\left(a+2b+3c\right)\)

\(\ge8+\dfrac{5}{4}.20=33\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\dfrac{3a}{4}=\dfrac{3}{a}\\\dfrac{b}{2}=\dfrac{9}{2b}\\\dfrac{c}{4}=\dfrac{4}{c}\\a+2b+3c=20\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\\c=4\end{matrix}\right.\)

Vậy \(MinA=33\), đạt được khi \(a=2;b=3;c=4\)

not biết làm

Tìm giá trị nhỏ nhất của S a b c a 2b c Giải: Dự đoán a=2,b=3,c=4 12 18 ... 18 16 4 S 4a 4b 4c a 2b 3c 3a 2b c ... 3 xy yz zx x2 y2 z2 Bài 11 Cho x, y là hai số thực không âm thay đổi. ..... 2 2 Bài 36 Cho a,b,c là các số thuộc 1; 2 thỏa mãn điều kiện a2+b2+c2 = 6.

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)