1.

\(\left(x+y\right)^2=\left(\dfrac{1}{2}.2x+\dfrac{1}{3}.3y\right)^2\le\left(\dfrac{1}{4}+\dfrac{1}{9}\right)\left(4x^2+9y^2\right)=\dfrac{169}{36}\)

\(\Rightarrow-\dfrac{13}{6}\le x+y\le\dfrac{13}{6}\)

Dấu "=" lần lượt xảy ra tại \(\left(-\dfrac{3}{2};-\dfrac{2}{3}\right)\) và \(\left(\dfrac{3}{2};\dfrac{2}{3}\right)\)

2.

\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y+\left(-\dfrac{1}{3}\right).6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)

\(\Rightarrow\left|y-2x\right|\le\dfrac{5}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\mp\dfrac{2}{5};\pm\dfrac{9}{20}\right)\)

3.

\(B^2=\left(6.\sqrt{x-1}+8\sqrt{3-x}\right)^2\le\left(6^2+8^2\right)\left(x-1+3-x\right)=200\)

\(\Rightarrow B\le2\sqrt{10}\)

Dấu "=" xảy ra khi \(\dfrac{\sqrt{x-1}}{6}=\dfrac{\sqrt{3-x}}{8}\Leftrightarrow x=\dfrac{43}{25}\)

\(B=6\sqrt{x-1}+6\sqrt{3-x}+2\sqrt{3-x}\ge6\sqrt{x-1}+6\sqrt{3-x}\)

\(B\ge6\left(\sqrt{x-1}+\sqrt{3-x}\right)\ge6\sqrt{x-1+3-x}=6\sqrt{2}\)

\(B_{min}=6\sqrt{2}\) khi \(\sqrt{3-x}=0\Rightarrow x=3\)

4.

\(49=\left(3a+4b\right)^2=\left(\sqrt{3}.\sqrt{3}a+2.2b\right)^2\le\left(3+4\right)\left(3a^2+4b^2\right)\)

\(\Rightarrow3a^2+4b^2\ge\dfrac{49}{7}=7\)

Dấu "=" xảy ra khi \(a=b=1\)

5.

\(\left(y-2x\right)^2=\left(\dfrac{1}{4}.4y-\dfrac{1}{3}.6x\right)^2\le\left(\dfrac{1}{16}+\dfrac{1}{9}\right)\left(16y^2+36x^2\right)=\dfrac{25}{16}\)

\(\Rightarrow-\dfrac{5}{4}\le y-2x\le\dfrac{5}{4}\)

\(\Rightarrow-\dfrac{5}{4}+5\le y-2x+5\le\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{15}{4}\le y-2x+5\le\dfrac{25}{4}\)

\(C_{min}=\dfrac{15}{4}\) khi \(\left(x;y\right)=\left(\dfrac{2}{5};-\dfrac{9}{20}\right)\)

\(C_{max}=\dfrac{25}{4}\) khi \(\left(x;y\right)=\left(-\dfrac{2}{5};\dfrac{9}{20}\right)\)

Lần sau lưu ý đăng câu hỏi 1 lần thôi, em đăng nhiều lần lặp lại sẽ bị xóa + ko ai giải cho đâu

\(x^3+y^3+y^3\ge3\sqrt[3]{x^3.y^3.y^3}=3xy^2\)

\(x^3+1+1\ge3x\)

\(2\left(y^3+1+1\right)\ge6y\)

Cộng vế:

\(2\left(x^3+2y^3\right)+6\ge3\left(x+2y+xy^2\right)=12\)

\(\Rightarrow x^3+2y^3\ge3\) (đpcm)

Dấu "=" xảy ra khi \(x=y=1\)

\(8y+4y+2y+y-5y=28,6\)

\(\Rightarrow9y=28,6\)

\(\Rightarrow y=\frac{286}{90}\)

y x 8 + y : 0,25 + y : 0,5 + y - y : 0,2 = 28,6 

=> y x 8 + y x 4 + y x 2 + y x 1 - y x 5 = 28,6

=> y x (8 + 4 + 2 + 1 - 5) = 28,6

=> y x 10 = 28,6

=> y = 28,6 : 10

=> y = 2,86

~Study well~

có chắc là đúng ko bạn ??? mk thấy sai sai 

a) y x 5 = 1948 + 247

y x 5 = 2195

y = 2195 : 5

y = 439

b) y : 3 = 190 - 90

y : 3 = 100

y = 100 x 3

y = 300

c) y - 8357 = 3829 x 2

y - 8357 = 7658

y = 7658 + 8357

y = 16 015

d) y x 8 = 182 x 4

y x 8 = 728

y = 728 : 8

y = 91

a) y x 5 = 1948 + 247

y x 5 = 2195

y = 2195 :5

y = 439

b) y : 3 = 190 - 90

y : 3 = 100

y = 100 x 3

y = 300

c) y - 8357 = 3829 x 2

y - 8357 = 7658

y = 7658 + 8357

y = 16015

d) y x 8 = 182 x 4

y x 8 = 728

y = 728 : 8

y = 91

a) y x 5 = 1948 + 247

y x 5 = 2195

y = 2195 : 5

y = 439

b) y : 3 = 190 - 90

y : 3 = 100

y = 100 x 3

y = 300

c) y - 8357 = 3829 x 2

y - 8357 = 7658

y = 7658 + 8357

y = 16 015

d) y x 8 = 182 x 4

y x 8 = 728

y = 728 : 8

y = 91

5rxdjexjgntrujnxgr6jexs6ue6thfydjytudcjxtyu45yuej8tuxr5ts

hellp

Em chiu thua luon

Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9\)

\(=x^4-3x^3-x^3+3x^2-5x^2+15x-3x+9\)

\(=\left(x-3\right)\left(x^3-x^2-5x-3\right)\)

\(=\left(x-3\right)\left(x^3-3x^2+2x^2-6x+x-3\right)\)

\(=\left(x-3\right)^2\cdot\left(x+1\right)^2\)

\(=\left(x^2-2x-3\right)^2\)

Bài 3: 

\(B=x^4-4x^3-2x^2+12x+9=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(3x^2+3x\right)+\left(9x+9\right)=\left(x^3-5x^2+3x+9\right)\left(x+1\right)=\left[\left(x^3+x^2\right)-\left(6x^2+6x\right)+\left(9x+9\right)\right]\left(x+1\right)=\left(x^2-6x+9\right)\left(x+1\right)^2=\left(x-3\right)^2\left(x+1\right)^2=\left[\left(x-3\right)\left(x+1\right)\right]^2\)

toán 9 mà sao câu 2a có giai thừa vậy bạn ??

chờ mình xíu nhé

a: a=-8

b: Khi x=-4 thì y=-8/-4=2

Khi x=8 thì y=-8/8=-1

\(x^2+2y^2-2xy+4y+3< 0\)

\(\Rightarrow x^2-2xy+y^2+y^2+4y+4-1< 0\)  

\(\Rightarrow\left(x^2-2xy+y^2\right)+\left(y^2+4y+4\right)-1< 0\)

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Mà: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x,y\\\left(y+2\right)^2\ge0\forall y\end{matrix}\right.\) 

\(\Rightarrow\left(x-y\right)^2+\left(y+2\right)^2-1\ge-1\forall x,y\)

Mặt khác: \(\left(x-y\right)^2+\left(y+2\right)^2-1< 0\)

Dấu "=" xảy ra:

\(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\)

\(\Rightarrow x=y=-2\)

Vậy: ....