a, Với \(x\ge0,x\ne4\)

\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)

\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

b, Ta có  \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)

\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)

1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

2 , A = \(1-\sqrt{2}\)

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

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

  \(\Leftrightarrow\)(\(\sqrt{x+1}-\sqrt{3x-1}\))(1-\(\sqrt{3x-1}\))=0

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

\(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)(Đk x≥\(\dfrac{1}{3}\))

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

=\(3x-1-x\sqrt{3x-1}+1\)

=\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

=\(\sqrt{\left(x+1\right)\left(3x-1\right)}-x\sqrt{x+1}+1\)

=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

ta có \(x\left(3-\sqrt{3x-1}\right)=\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)=\sqrt{x+1}\left(\sqrt{3x-1}-x\right)\)

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

  Chúc Bạn Học Tốt.

b, \(\frac{a^3}{b+2c}+\frac{b^3}{c+2a}+\frac{c^3}{a+2b}\ge1\)

\(\frac{a^4}{ab+2ac}+\frac{b^4}{bc+2ab}+\frac{c^4}{ac+2bc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)( Bunhia dạng phân thức )

mà \(a^2+b^2+c^2\ge ab+bc+ac\)

\(=\frac{\left(ab+bc+ac\right)^2}{3+2\left(ab+ac+bc\right)}=\frac{9}{3+6}=1\)( đpcm ) 

1.

Điều kiện x \ge \dfrac14x≥41​.

Phương trình tương đương với \left(\sqrt2.\sqrt{2x^2+x+1}-2\right)-\left(\sqrt{4x-1}-1\right)+2x^2+3x-2 = 0(2​.2x2+x+1​−2)−(4x−1​−1)+2x2+3x−2=0 \Leftrightarrow \dfrac{4x^2+2x-2}{\sqrt2.\sqrt{2x^2+x+1}+2} - \dfrac{4x-2}{\sqrt{4x-1}+1} + (x+2)(2x-1) = 0⇔2​.2x2+x+1​+24x2+2x−2​−4x−1​+14x−2​+(x+2)(2x−1)=0\\ \Leftrightarrow (2x-1)\left(\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2\right) = 0⇔(2x−1)(2​2x2+x+1​+22(x+1)​−4x−1​+12​+x+2)=0

\Leftrightarrow \left[\begin{aligned} & x =\dfrac12\\ & \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 = 0\\ \end{aligned}\right.⇔⎣⎢⎢⎢⎡​​x=21​2​2x2+x+1​+22(x+1)​−4x−1​+12​+x+2=0​

Với x \ge \dfrac14x≥41​ ta có:

\dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} > 02​2x2+x+1​+22(x+1)​>0

- \dfrac2{\sqrt{4x-1}+1} \ge -2−4x−1​+12​≥−2

$x+2>2$.

Suy ra \dfrac{2(x+1)}{\sqrt2 \sqrt{2x^2+x+1}+2} - \dfrac2{\sqrt{4x-1}+1} + x + 2 > 02​2x2+x+1​+22(x+1)​−4x−1​+12​+x+2>0.

Vậy phương trình có nghiệm duy nhất x = \dfrac12.x=21​.

2.

Đặt P = \dfrac{a^3}{b+2c} + \dfrac{b^3}{c+2a} + \dfrac{c^3}{a+2b}P=b+2ca3​+c+2ab3​+a+2bc3​

Áp dụng bất đẳng thức Cauchy cho hai số dương \dfrac{9a^3}{b + 2c}b+2c9a3​ và $(b+2c)a$ ta có

\dfrac{9a^3}{b+2c} + (b+2c)a \ge 6a^2b+2c9a3​+(b+2c)a≥6a2.

Tương tự \dfrac{9b^3}{c+2a} + (c+2a)b \ge 6b^2c+2a9b3​+(c+2a)b≥6b2, \dfrac{9c^3}{a+2b} + (a+2b)c \ge 6c^2a+2b9c3​+(a+2b)c≥6c2.

Cộng các vế ta có 9P + 3(ab+bc+ca) \ge 6(a^2+b^2+c^2)9P+3(ab+bc+ca)≥6(a2+b2+c2).

Mà a^2+b^2+c^2 \ge ab+bc+ca = 4a2+b2+c2≥ab+bc+ca=4 nên $P\ge 1$ (ta có đpcm).

1.

√2 × √(2x²+x+1)        +      √(4x-1) + 3x-3=0

⇌[√(4x²+2x+2)-2] - [√(4x-1)     -1] + (2x²+3x-2)=0

⇌(4x²+2x-2)/[√(4x²+2x+2)+2] - (4x-2)/[√(4x-1)+1] + (2x-1)(x+2) =0

⇔(2x-1) × [(2x+2)/√(4x²+2x+2+2) - 2/(√4x-1)+1+x+2]=0

Với x≥1/4 thì (2x+2)/(√4x²+2x+2+2)≥0 hoặc x+2>2 hoặc (√4x-1)+1≥1 ⇌ 2/[(√4x-1)+1]≤2

⇒(2x+2)/[(√4x²+2x+2)+2] - 2/[(x-1)+1]+x+2>0-2+2=0

⇌ 2x-1=0⇒x=1/2 

Vậy x=1/2

2.

Áp dụng bất đẳng thức ta có :

Vế trái = a⁴/(ab +2ac)    +   b⁴/(bc+2ab)  + c⁴/(ac+2bc)≥[(a² + b² +c²)²]/[3(ab+bc+ca) =[(a²+b²+c²)²]/9

Ấp dụng bất đẳng thức ta có :

ab+bc+ca≤a²+b²+c² 

Vế trái ≥ [(a²+b²+c²)]/9≥3²/9 =1

⇒ Vế trái ≥1 (đpcm)

Dấu = xảy ra khi a=b=c=1

Cậu sống ở đâu hở ? Lấy đâu ra toán khó thế ?

Câu 3 sửa \(\int\limits_1^{3/2} \)

Khó như thế này thì ai mà làm nổi

a: \(\left(2\sqrt{10}+3\sqrt{3}\right)^2=67+12\sqrt{30}\)

\(\left(3\sqrt{5}+2\sqrt{7}\right)^2=77+12\sqrt{35}\)

mà \(12\sqrt{30}< 12\sqrt{35};67< 77\)

nên \(2\sqrt{10}+3\sqrt{3}< 3\sqrt{5}+2\sqrt{7}\)

b: \(\left(\sqrt{2}+\sqrt{3}\right)^2=5+2\sqrt{6}\)

\(2^2=4\)

mà 5>4

nên \(\sqrt{2}+\sqrt{3}>2\)

\(\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6+\sqrt{2}}\right)}=2\)

=2.

1 bài Mincopxki khá quen:

\(P\ge\sqrt{\left(a+b+c\right)^2+\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2}\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{\left(a+b+c\right)^2}}\)

Đến đây thì nó là bài Cô-si có biên, cứ tách ghép theo điểm rơi là được:

\(P\ge\sqrt{\left(a+b+c\right)^2+\dfrac{81}{16\left(a+b+c\right)^2}+\dfrac{1215}{16\left(a+b+c\right)^2}}\)

\(P\ge\sqrt{2\sqrt{\dfrac{81\left(a+b+c\right)^2}{16\left(a+b+c\right)^2}}+\dfrac{1215}{16.\left(\dfrac{3}{2}\right)^2}}=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xayr a khi \(a=b=c=\dfrac{1}{2}\)

Áp dụng bđt AM - GM ta có \(\sqrt{\dfrac{a^2+\left(b+c\right)^2}{2a\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a^2+\left(b+c\right)^2}{2a\left(b+c\right)}+1\right)=\dfrac{1}{2}\dfrac{\left(a+b+c\right)^2}{2a\left(b+c\right)}\)

\(\Rightarrow\sqrt{\dfrac{a\left(b+c\right)}{a^2+\left(b+c\right)^2}}\ge\dfrac{2\sqrt{2}a\left(b+c\right)}{\left(a+b+c\right)^2}\).

Tương tự,...

Cộng vế với vế ta có \(\sqrt{\dfrac{a\left(b+c\right)}{a^2+\left(b+c\right)^2}}+\sqrt{\dfrac{b\left(c+a\right)}{b^2+\left(c+a\right)^2}}+\sqrt{\dfrac{c\left(a+b\right)}{c^2+\left(a+b\right)^2}}\ge\dfrac{4\sqrt{2}\left(ab+bc+ca\right)}{\left(a+b+c\right)^2}\). (*)

Mặt khác do a, b, c là độ dài ba cạnh của 1 tam giác nên \(a\left(b+c-a\right)+b\left(c+a-b\right)+c\left(a+b-c\right)>0\Rightarrow2\left(ab+bc+ca\right)\ge a^2+b^2+c^2\Rightarrow4\left(ab+bc+ca\right)\ge\left(a+b+c\right)^2\). (**)

Từ (*) và (**) ta có đpcm.