đây nhé bn

tick cho mk nha

ảo ma à??

dở à bn gì ơi

wibu chính hiệu

Câu 285

a) ĐKXĐ: $x\le 10.$

 \(PT\Leftrightarrow\left(\dfrac{x^3+7x^2+18x+4}{\sqrt{10-x}}-10\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\dfrac{\left(x^5+15x^4+100x^3+360x^2+740x+984\right)}{\sqrt{10-x}\left(x^3+7x^2+8x+4+10\sqrt{10-x}\right)}+1\right]=0\)

Rõ ràng biểu thức trong ngoặc vuông vô nghiệm.

Vậy $x=1$ (TMĐKXĐ)

b) Đặt $t=ab+bc+ca.$

 \(a,b,c\in\left[0,1\right]\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1.\) (1)

Từ (1) suy ra \(3abc\ge\sum c\left(a+b-1\right)=2t-\left(a+b+c\right)\ge2t-3\)

Cũng do $a,b,c\in \left[0,1\right]$ suy ra \(\left(a-1\right)\left(b-1\right)\left(c-1\right)\le0\Rightarrow abc\le\sum\left(ab-a\right)+1\)

Do đó"\(VT\le\sum\dfrac{a}{1+bc}+\sum\left(ab-a\right)+1\)

\(=\sum\left(\dfrac{a}{1+bc}-a\right)+\sum ab+1\)

\(=-abc\sum\dfrac{1}{1+bc}+ab+bc+ca+1\)

\(\le t+1-\dfrac{9abc}{t+3}\le t+1-\dfrac{3\left(2t-3\right)}{t+3}\le\dfrac{5}{2}\) 

\(\Leftrightarrow\left(2t-3\right)\left(3-t\right)\ge0\)

Do \(t\le\dfrac{\left(a+b+c\right)^2}{3}=3\) nên nếu $ab+bc+ca\ge \dfrac{3}{2}$ thì bất đẳng thức đúng.

Trong trường hợp ngược lại ta có \(VT\le t+1-\dfrac{9abc}{t+3}\le t+1\le\dfrac{3}{2}+1=\dfrac{5}{2}\) (đpcm)

Hoàn tất chứng minh.

Đẳng thức xảy ra khi (bạn đọc tự xét)

290

Ta có \(\dfrac{a^4b}{a^2+1}=a^2b-\dfrac{a^2b}{a^2+1}\ge a^2b-\dfrac{a^2b}{2a}=a^2b-\dfrac{ab}{2}\)

Chứng minh tương tự ta được:  

\(\dfrac{b^4c}{b^2+1}\ge b^2c-\dfrac{bc}{2};\dfrac{c^4a}{c^2+1}\ge c^2a-\dfrac{ca}{2}\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{ab}{2}-\dfrac{bc}{2}-\dfrac{ca}{2}\)

Áp dụng bđt Cô-si:

\(a^2b+a^2b+b^2c\ge3\sqrt[3]{a^2b\cdot a^2b\cdot b^2c}=3\sqrt[3]{a^3b^3\cdot abc}=3ab\)

Tương tự: \(b^2c+b^2c+c^2a\ge3bc;c^2a+c^2a+a^2b\ge3ca\)

\(\Rightarrow a^2b+a^2b+b^2c+b^2c+b^2c+c^2a+c^2a+c^2a+a^2b\ge3ab+3bc+3ca\Rightarrow3\left(a^2b+b^2c+c^2a\right)\ge3\left(ab+bc+ca\right)\Rightarrow a^2b+b^2c+c^2a\ge ab+bc+ca\)

\(\Rightarrow\dfrac{a^4b}{a^2+1}+\dfrac{b^4c}{b^2+1}+\dfrac{c^4a}{c^2+1}\ge a^2b+b^2c+c^2a-\dfrac{1}{2}\left(ab+bc+ca\right)\ge ab+bc+ca-\dfrac{1}{2}\left(ab+bc+ca\right)=\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{3}{2}\sqrt[3]{\left(abc\right)^2}=\dfrac{3}{2}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

C290: [Toán] - Hoc24

Chúc mừng mọi người (CTV 1)

Bài nào đó k ghi số nên không bt gọi ntn:

Chuẩn hóa x + y + z = 3. Ta cần cm \(x^2y+y^2z+z^2x+xyz\le4\).

Giả sử \(z=mid\left\{x,y,z\right\}\Rightarrow\left(x-z\right)\left(y-z\right)\le0\)

\(\Leftrightarrow xy+z^2\le xz+yz\)

\(\Leftrightarrow x^2y+xz^2\le x^2z+xyz\).

Từ đó \(x^2y+y^2z+z^2x+xyz\le x^2z+xyz+y^2z+xyz=z\left(x+y\right)^2\le\dfrac{\dfrac{\left(2z+x+y+x+y\right)^3}{27}}{2}=4\).

Câu cuối:

Áp dụng BĐT BSC:

\(\dfrac{a}{\sqrt{a^2+b+c}}=\sqrt{\dfrac{a^2}{a^2+b+c}}=\sqrt{\dfrac{a^2\left(1+b+c\right)}{\left(a^2+b+c\right)\left(1+b+c\right)}}\le\sqrt{\dfrac{a^2\left(1+b+c\right)}{\left(a+b+c\right)^2}}\le\dfrac{a\sqrt{1+b+c}}{a+b+c}\)

Tương tự \(\dfrac{b}{\sqrt{b^2+c+a}}=\le\dfrac{b\sqrt{1+c+a}}{a+b+c}\); \(\dfrac{c}{\sqrt{c^2+a+b}}=\le\dfrac{c\sqrt{1+a+b}}{a+b+c}\)

Khi đó \(VT\le\Sigma\left(\dfrac{a}{a+b+c}.\sqrt{1+b+c}\right)\)

Giả sử \(a\ge b\ge c\)

Áp dụng BĐT Chebyshev với bộ \(\dfrac{a}{a+b+c};\dfrac{b}{a+b+c};\dfrac{c}{a+b+c}\) và \(\sqrt{1+b+c};\sqrt{1+c+a};\sqrt{1+a+b}\):

\(VT\le\dfrac{1}{3}\Sigma\dfrac{a}{a+b+c}.\Sigma\sqrt{1+a+b}=\dfrac{\Sigma\sqrt{1+a+b}}{3}\)

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

\(\le\dfrac{\sqrt{9+6\sqrt{3\left(a^2+b^2+c^2\right)}}}{3}=\sqrt{3}\)

Đẳng thức xảy ra khi \(a=b=c=1\)

Bài 1 GPT: \(x^2+2018\sqrt{2x^2+1}=x+1+2018\sqrt{x^2+x+1}\)(1) ĐKXĐ: \(\forall x\in R\)

(1) \(\Leftrightarrow x^2-x-1+2018\sqrt{2x^2+1}-2018\sqrt{x^2+x+1}=0\)

\(\Rightarrow x^2-x-1+2018\cdot\dfrac{\left(\sqrt{2x^2+1}-\sqrt{x^2+x+2}\right)\left(\sqrt{2x^2+1}+\sqrt{x^2+x+2}\right)}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}=0\)

\(\Leftrightarrow x^2-x-1+2018\cdot\dfrac{\left(x^2-x-1\right)}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}=0\)

\(\Leftrightarrow\left(x^2-x-1\right)\left(1+\dfrac{2018}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}\right)=0\)

\(\Leftrightarrow x^2-x-1=0\) vì \(1+\dfrac{2018}{\sqrt{2x^2+1}+\sqrt{x^2+x+2}}>1>0\forall x\)

\(\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\) Vậy...

C280:

Áp dụng BĐT AM-GM và BĐT BSC:

\(\dfrac{1}{\sqrt{x+3y}}+\sqrt{x+3y}\ge2\Rightarrow\dfrac{1}{\sqrt{x+3y}}\ge2-\sqrt{x+3y}\)

\(\dfrac{1}{\sqrt{y+3z}}+\sqrt{y+3z}\ge2\Rightarrow\dfrac{1}{\sqrt{y+3z}}\ge2-\sqrt{y+3z}\)

\(\dfrac{1}{\sqrt{z+3x}}+\sqrt{z+3x}\ge2\Rightarrow\dfrac{1}{\sqrt{z+3x}}\ge2-\sqrt{z+3x}\)

\(\Rightarrow P=\dfrac{1}{\sqrt{x+3y}}+\dfrac{1}{\sqrt{y+3z}}+\dfrac{1}{\sqrt{z+3x}}\)

\(\ge6-\left(\sqrt{x+3y}+\sqrt{y+3z}+\sqrt{z+3x}\right)\)

\(\ge6-\sqrt{3\left(x+3y+y+3z+z+3x\right)}\)

\(=6-\sqrt{12\left(x+y+z\right)}=3\)

\(minP=3\Leftrightarrow a=b=c=\dfrac{1}{4}\)

Bài 7) 

\(bđt\Leftrightarrow4\left(a+b+c\right)\left(a^2+b^2+c^2\right)-3\left(a^3+b^3+c^3\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow a^3+b^3+c^3+4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow4ab\left(a+b\right)+4bc\left(b+c\right)+4ac\left(a+c\right)\ge3ab\left(a+b\right)+3bc\left(b+c\right)+3ac\left(a+c\right)+6abc\)\(\Leftrightarrow ab\left(a+b\right)+bc\left(b+c\right)+ac\left(a+c\right)\ge6abc\)

\(\Leftrightarrow\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\ge6\)

(Đúng theo Cô Si)

"=" khi a=b=c=1

281:

Ta có:\(ab+bc+ca=3abc\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=3\)

\(\dfrac{1}{\sqrt{a^3+b}}\le\dfrac{1}{\sqrt{2\sqrt{a^3b}}}=\dfrac{1}{\sqrt{2a}\cdot\sqrt[4]{ab}}\le\dfrac{1}{2\sqrt{2a}}\cdot\left(\dfrac{1}{\sqrt{a}}+\dfrac{1}{\sqrt{b}}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\right)\le\dfrac{1}{2\sqrt{2}}\cdot\left[\dfrac{1}{a}+\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\right]=\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}\right)\) Chứng minh tương tự:

\(\dfrac{1}{\sqrt{b^3+c}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}\right);\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\cdot\left(\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)\)\(\Rightarrow\dfrac{1}{\sqrt{a^3+b}}+\dfrac{1}{\sqrt{b^3+c}}+\dfrac{1}{\sqrt{c^3+a}}\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{1}{a}+\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{b}+\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{c}+\dfrac{1}{2c}+\dfrac{1}{2a}\right)=\dfrac{1}{2\sqrt{2}}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{3}{\sqrt{2}}\) Dấu = xảy ra \(\Leftrightarrow a=b=c=1\)

Câu 5 em thấy thầy làm từ chiều, em nghĩ anh nên đổi câu khác:

Cho \(x,y,z\ge0\).Tìm giá trị lớn nhất :\(P=\dfrac{x}{x^2 y^2 2} \dfrac{y}{y^2 z^2 2} \dfrac{z}{z^2 x^2 2}\) - Hoc24

Câu 266 là >= chứ nhỉ?

Câu 5 (có chữ HẾT (.❛ ᴗ ❛.) )

Đặt \(P=a\sqrt{b^3+1}+b\sqrt{c^3+1}+c\sqrt{a^3+1}\)

Ta có:

\(a\ge0\Rightarrow b^3+1\ge1\Rightarrow a\sqrt{b^3+1}\ge a\)

Hoàn toàn tương tự, ta có: \(\left\{{}\begin{matrix}b\sqrt{c^3+1}\ge b\\c\sqrt{a^3+1}\ge c\end{matrix}\right.\)

Cộng vế: \(P\ge a+b+c=3\) (đpcm)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;0;3\right)\) và các hoán vị

\(a\sqrt{b^3+1}=a\sqrt{\left(b+1\right)\left(b^2-b+1\right)}\le\dfrac{1}{2}a\left(b^2+2\right)=\dfrac{1}{2}ab^2+a\)

Tương tự: \(b\sqrt{c^3+1}\le\dfrac{1}{2}bc^2+b\) ; \(c\sqrt{a^3+1}\le\dfrac{1}{2}ca^2+c\)

Cộng vế: \(P\le\dfrac{1}{2}\left(ab^2+bc^2+ca^2\right)+3\)

Không mất tính tổng quát, giả sử \(a=mid\left\{a;b;c\right\}\)

\(\Rightarrow\left(a-b\right)\left(a-c\right)\le0\Leftrightarrow a^2+bc\le ac+ab\Rightarrow ca^2+bc^2\le ac^2+abc\)

\(\Rightarrow ab^2+bc^2+ca^2\le ab^2+ac^2+abc\le ab^2+ac^2+2abc=a\left(b+c\right)^2\)

\(\Rightarrow ab^2+bc^2+ca^2\le\dfrac{1}{2}.2a\left(b+c\right)\left(b+c\right)\le\dfrac{1}{54}\left(2a+2b+2c\right)^3=4\)

\(\Rightarrow P\le\dfrac{1}{2}.4+3=5\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;2;0\right)\) và 1 số hoán vị

C113

Ta có: \(\dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c} = \dfrac{1}{{a + b + c}} \Longrightarrow \dfrac{1}{a} + \dfrac{1}{b} = \dfrac{1}{{a + b + c}} - \dfrac{1}{c}\)

\(\begin{array}{l} \Longrightarrow \left( {a + b} \right)\left( {a + b + c} \right)c = abc - ab\left( {a + b + c} \right)\\ \Longrightarrow \left( {a + b} \right)\left( {b + c} \right)\left( {c + a} \right) = 0 \end{array}\)

............

C112:

a¹⁶ + a⁸b⁸ + b¹⁶ 

= a¹⁶ + 2a⁸b⁸ + b¹⁶ - a⁸b⁸

= (a⁸ + b⁸)² - (a⁴b⁴)²

= (a⁸ + b⁸ - a⁴b⁴)(a⁸ + b⁸ + a⁴b⁴)

C112 : \(a^{16}+a^8b^8+b^{16}\)

\(=\left(a^8+b^8\right)^2-\left(a^4b^4\right)^2\)

\(=\left(a^8+b^8+a^4b^4\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left[\left(a^4+b^4\right)^2-\left(a^2b^2\right)^2\right].\left(a^8+b^8-a^4b^4\right)\)

\(=\left(a^4+b^4-a^2b^2\right)\left(a^4+b^4+a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left[\left(a^2+b^2\right)^2-\left(ab\right)^2\right]\left(a^4+b^4-a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

\(=\left(a^2+b^2+ab\right)\left(a^2+b^2-ab\right)\left(a^4+b^4-a^2b^2\right)\left(a^8+b^8-a^4b^4\right)\)

Bài II:

1) \(PT\Leftrightarrow3x^2+2y^2+z^2+4xy+2yz+2zx=26\)

\(\Leftrightarrow\left(x+y+z\right)^2+\left(x+y\right)^2+x^2=26\).

Tách \(26=0^2+1^2+5^2=1^2+3^2+4^2\).

Mặt khác ta có x + y + z > x + y > x > 0 nên ta phải có x = 1; x + y = 3; x + y + z = 4.

Từ đó x = 1; y = 2; z = 1.

Vậy nghiệm nguyên dương của phương trình là (x, y, z) = (1; 2; 1).

Bài I :

1 ĐKXĐ \(x\ge\dfrac{-1}{8}\) 

\(\Leftrightarrow9x+17-6\sqrt{8x+1}-4\sqrt{x+3}=0\) 

\(\Leftrightarrow8x+1-6\sqrt{8x+1}+9+x+3-4\sqrt{x+3}+4=0\)

\(\Leftrightarrow\left(\sqrt{8x+1}-3\right)^2+\left(\sqrt{x+3}-2\right)^2=0\) 

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{8x+1}-3=0\\\sqrt{x+3}-2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{8x+1}=3\\\sqrt{x+3}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}8x+1=9\\x+3=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}8x=8\\x=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=1\end{matrix}\right.\)

\(\Leftrightarrow x=1\left(TM\right)\)

 Vậy...

đây là đề thi chuyên phải không ạ?