ĐKXĐ: \(0\le x\le1\)

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

\(P\ge\sqrt{1-x+x}+\sqrt{1+x}+\sqrt{x}\)

\(P\ge1+\sqrt{1+x}+\sqrt{x}\ge1+1+0=2\)

\(P_{min}=2\) khi \(x=0\)

1) Áp dụng bđt Cauchy cho 3 số dương ta có

 \(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}+x^3\ge4\sqrt[4]{\dfrac{1}{x}.\dfrac{1}{x}.\dfrac{1}{x}.x^3}=4\) (1)

\(\dfrac{3}{y^2}+y^2\ge2\sqrt{\dfrac{3}{y^2}.y^2}=2\sqrt{3}\) (2)

\(\dfrac{3}{z^3}+z=\dfrac{3}{z^3}+\dfrac{z}{3}+\dfrac{z}{3}+\dfrac{z}{3}\ge4\sqrt[4]{\dfrac{3}{z^3}.\dfrac{z}{3}.\dfrac{z}{3}.\dfrac{z}{3}}=4\sqrt{3}\) (3)

Cộng (1);(2);(3) theo vế ta được

\(\left(\dfrac{3}{x}+\dfrac{3}{y^2}+\dfrac{3}{z^3}\right)+\left(x^3+y^2+z\right)\ge4+2\sqrt{3}+4\sqrt{3}\)

\(\Leftrightarrow3\left(\dfrac{1}{x}+\dfrac{1}{y^2}+\dfrac{1}{z^3}\right)\ge3+4\sqrt{3}\)

\(\Leftrightarrow P\ge\dfrac{3+4\sqrt{3}}{3}\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{x}=x^3\\\dfrac{3}{y^2}=y^2\\\dfrac{3}{z^3}=\dfrac{z}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\sqrt[4]{3}\\z=\sqrt{3}\end{matrix}\right.\) (thỏa mãn giả thiết ban đầu)

2) Ta có \(4\sqrt{ab}=2.\sqrt{a}.2\sqrt{b}\le a+4b\)

Dấu"=" khi a = 4b

nên \(\dfrac{8}{7a+4b+4\sqrt{ab}}\ge\dfrac{8}{7a+4b+a+4b}=\dfrac{1}{a+b}\)

Khi đó \(P\ge\dfrac{1}{a+b}-\dfrac{1}{\sqrt{a+b}}+\sqrt{a+b}\)

Đặt \(\sqrt{a+b}=t>0\) ta được

\(P\ge\dfrac{1}{t^2}-\dfrac{1}{t}+t=\left(\dfrac{1}{t^2}-\dfrac{2}{t}+1\right)+\dfrac{1}{t}+t-1\)

\(=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\)

Có \(\dfrac{1}{t}+t\ge2\sqrt{\dfrac{1}{t}.t}=2\) (BĐT Cauchy cho 2 số dương)

nên \(P=\left(\dfrac{1}{t}-1\right)^2+\dfrac{1}{t}+t-1\ge\left(\dfrac{1}{t}-1\right)^2+1\ge1\)

Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\dfrac{1}{t}-1=0\\t=\dfrac{1}{t}\end{matrix}\right.\Leftrightarrow t=1\)(tm)

khi đó a + b = 1

mà a = 4b nên \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

Vậy MinP = 1 khi \(a=\dfrac{4}{5};b=\dfrac{1}{5}\)

4:

Sửa đề: Tìm giá trị lớn nhất

ĐKXĐ: x>=0

Đặt \(A=\frac{\sqrt{x}}{2\sqrt{x}+1}+\frac{1}{2\sqrt{x}+1}\)

\(=\frac{\sqrt{x}+1}{2\sqrt{x}+1}=\frac12\cdot\frac{2\sqrt{x}+2}{2\sqrt{x}+1}=\frac12\left(1+\frac{1}{2\sqrt{x}+1}\right)\)

Ta có: \(2\sqrt{x}+1\ge1\forall x\) thỏa mãn ĐKXĐ
=>\(\frac{1}{2\sqrt{x}+1}\le\frac11=1\forall x\) thỏa mãn ĐKXĐ

=>\(\frac{1}{2\sqrt{x}+1}+1\le2\forall x\) thỏa mãn ĐKXĐ

=>\(\frac12\left(1+\frac{1}{2\sqrt{x}+1}\right)\le2\cdot\frac12=1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

Để \(\frac{x^2+7}{x+1}\)nhận giá trị nguyên thì \(x^2+7⋮x+1\left(1\right)\)

+)Ta có:\(x+1⋮x+1\)

\(\Rightarrow x.\left(x+1\right)⋮x+1\)

\(\Rightarrow x^2+x⋮x+1\left(2\right)\)

+)Từ (1) và (2)

\(\Rightarrow\left(x^2+x\right)-\left(x^2+7\right)⋮x+1\)

\(\Rightarrow x^2+x-x^2-7⋮x+1\)

\(\Rightarrow x-7⋮x+1\left(3\right)\)

+)Ta lại có:\(x+1⋮x+1\left(4\right)\)

+)Từ (3) và (4)

\(\Rightarrow\left(x+1\right)-\left(x-7\right)⋮x+1\)

\(\Rightarrow x+1-x+7⋮x+1\)

\(\Rightarrow8⋮x+1\)

\(\Rightarrow x+1\in\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1;-5;3;-9;7\right\}\in Z\)

Vậy \(x\in\left\{-2;0;-3;1;-5;3;-9;7\right\}\)

Chúc bn học tốt

Ta có: \(A=\sqrt{x-1-2\sqrt{x-2}}-\sqrt{x+7-6\sqrt{x-2}}\)

\(=\sqrt{x-2-2\cdot\sqrt{x-2}\cdot1+1}-\sqrt{x-2-6\cdot\sqrt{x-2}+9}\)

\(=\sqrt{\left(\sqrt{x-2}-1\right)^2}-\sqrt{\left(\sqrt{x-2}-3\right)^2}=\left|\sqrt{x-2}-1\right|-\left|\sqrt{x-2}-3\right|\)

=>\(A\le\left|\sqrt{x-2}-1-\sqrt{x-2}+3\right|=2\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x-2}-1\right)\left(\sqrt{x-2}-3\right)\ge0\)

=>\(\left[\begin{array}{l}\sqrt{x-2}\ge3\\ \sqrt{x-2}\le1\end{array}\right.\Rightarrow\left[\begin{array}{l}x-2\ge9\\ 0\le x-2\le1\end{array}\right.\Rightarrow\left[\begin{array}{l}x\ge11\\ 2\le x\le3\end{array}\right.\)

\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

Biểu thức này không có GTNN. Nếu muốn bạn cần bổ sung thêm điều kiện.

Trả lời:

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

Ta có: \(x^2\ge0\forall x\)

\(\Rightarrow-x^2\le0\forall x\)

\(\Rightarrow1-x^2\le1\forall x\)

\(\Rightarrow\sqrt{1-x^2}\le1\forall x\)

\(\Rightarrow1+\sqrt{1-x^2}\le2\forall x\)

\(\Rightarrow\dfrac{1}{1+\sqrt{1-x^2}}\ge\dfrac{1}{2}\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy GTNN của biểu thức là 1/2 <=> x = 0