Chương 4: BẤT ĐẲNG THỨC, BẤT PHƯƠNG TRÌNH

\(a-b+\dfrac{4}{b\left(a-b\right)}+b\ge3\sqrt[3]{\dfrac{4b\left(a-b\right)}{a-b}}=3\sqrt[3]{4}>3\)

Dấu "=" không xảy ra 

1.

a, ĐK: \(x\ge-1\)

\(\Leftrightarrow x+\sqrt{x+1}< 2+\sqrt{x+1}\)

\(\Leftrightarrow x< 2\)

\(\Rightarrow x\in[-1;2)\)

b, ĐK: \(x< \dfrac{1}{2}\)

\(\dfrac{2-x}{\sqrt{1-2x}}\le\dfrac{1}{\sqrt{1-2x}}\)

\(\Leftrightarrow2-x\le1\)

\(\Leftrightarrow x\ge1\)

\(\Rightarrow\) vô nghiệm

c, ĐK: \(x\le3\)

\(\left(x-1\right)\sqrt{3-x}\ge0\)

\(\Leftrightarrow x-1\ge0\)

\(\Leftrightarrow x\ge1\)

\(\Rightarrow x\in\left[1;3\right]\)

2.

ĐK: \(x\le1\)

\(\left(x-m\right)\sqrt{1-x}\ge0\)

\(\Leftrightarrow x-m\ge0\)

\(\Leftrightarrow m\le x\le1\)

Vậy \(m\le1\)

5.

\(2x-m< 3\left(x-1\right)\)

\(\Leftrightarrow x>3-m\)

Yêu cầu bài toán thỏa mãn khi \(3-m=4\Leftrightarrow m=-1\)

6. 

\(m\left(2x-1\right)\ge2x+1\)

\(\Leftrightarrow2\left(m-1\right)x\ge m+1\left(1\right)\)

TH1: \(m< 1\)

\(\left(1\right)\Leftrightarrow x\le\dfrac{m+1}{2m-2}\)

\(\Rightarrow\) Không tồn tại giá trị m thỏa mãn

TH2: \(m=1\)

\(\Leftrightarrow0x\ge2\left(vn\right)\)

TH3: \(m>1\)

\(\left(1\right)\Leftrightarrow x\ge\dfrac{m+1}{2m-2}\)

Yêu câu bài toán thỏa mãn khi \(\dfrac{m+1}{2m-2}=1\Leftrightarrow m=3\left(tm\right)\)

Vậy \(m=3\)

Chứng minh: \(x^3+y^3\ge xy\left(x+y\right)\left(1\right)\)

\(x^3+y^3\ge xy\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)\ge xy\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^3\ge4xy\left(x+y\right)\)

\(\Leftrightarrow\left(x+y\right)^2\ge4xy\)

\(\Leftrightarrow\left(x-y\right)^2\ge0\) đúng

\(\Rightarrow\left(1\right)\) đúng

Áp dụng BĐT \(x^3+y^3\ge xy\left(x+y\right)\)

\(\dfrac{a^3+b^3}{ab}+\dfrac{b^3+c^3}{bc}+\dfrac{c^3+a^3}{ca}\)

\(\ge\dfrac{ab\left(a+b\right)}{ab}+\dfrac{bc\left(b+c\right)}{bc}+\dfrac{ca\left(c+a\right)}{ca}\)

\(=2\left(a+b+c\right)\)

a, \(\left|x+2\right|+\left|-2x+1\right|\le x+1\left(1\right)\)

TH1: \(x\le-2\)

\(\Rightarrow x+1\le-1< \left|x+2\right|+\left|-2x+1\right|\)

\(\Rightarrow\) vô nghiệm

TH2: \(-2< x\le\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow x+2-2x+1\le x+1\)

\(\Leftrightarrow x\ge1\)

\(\Rightarrow x\in\left[1;\dfrac{1}{2}\right]\)

TH3: \(x>\dfrac{1}{2}\)

\(\left(1\right)\Leftrightarrow x+2+2x-1\le x+1\)

\(\Leftrightarrow x\le0\)

\(\Rightarrow\) vô nghiệm

Vậy \(x\in\left[1;\dfrac{1}{2}\right]\)

b, \(\left|x+2\right|-\left|x-1\right|< x-\dfrac{3}{2}\left(2\right)\)

TH1: \(x\le-2\)

\(\left(2\right)\Leftrightarrow-x-2+x-1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x>-\dfrac{3}{2}\)

\(\Rightarrow\) vô nghiệm

TH2: \(-2< x\le1\)

\(\left(2\right)\Leftrightarrow x+2+x-1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x< -\dfrac{5}{2}\)

\(\Rightarrow\) vô nghiệm

TH3: \(x>1\)

\(\left(2\right)\Leftrightarrow x+2-x+1< x-\dfrac{3}{2}\)

\(\Leftrightarrow x>\dfrac{9}{2}\)

\(\Rightarrow x\in\left(\dfrac{9}{2};+\infty\right)\)

Vậy \(x\in\left(\dfrac{9}{2};+\infty\right)\)

c, Tương tự a,b

d, ĐK: \(x\ne-2;x\ne1\)

\(\left|\dfrac{-5}{x+2}\right|< \left|\dfrac{10}{x-1}\right|\)

\(\Leftrightarrow\dfrac{1}{\left|x+2\right|}< \dfrac{2}{\left|x-1\right|}\)

\(\Leftrightarrow2\left|x+2\right|>\left|x-1\right|\)

\(\Leftrightarrow4\left(x+2\right)^2>\left(x-1\right)^2\)

\(\Leftrightarrow4\left(x^2+4x+4\right)>x^2-2x+1\)

\(\Leftrightarrow3x^2+18x+15>0\)

\(\Leftrightarrow...\)

e, ĐK: \(x\ne-1\)

\(\left|\dfrac{2-3\left|x\right|}{1+x}\right|\le1\)

\(\Leftrightarrow\left|2-3\left|x\right|\right|\le\left|x+1\right|\)

\(\Leftrightarrow\left(2-3\left|x\right|\right)^2\le\left(x+1\right)^2\)

\(\Leftrightarrow4+9x^2-12\left|x\right|\le x^2+2x+1\)

\(\Leftrightarrow8x^2-12\left|x\right|-2x+3\le0\)

Đến đây dễ rồi, xét hai trường hợp để phá dấu giá trị tuyệt đối rồi đối chiếu điêì kiện.

a, \(\left|4x-8\right|\le8\)

\(\Leftrightarrow\left(\left|4x-8\right|\right)^2\le64\)

\(\Leftrightarrow16x^2-64x+64\le64\)

\(\Leftrightarrow16x^2-64x\le0\)

\(\Leftrightarrow16x\left(x-4\right)\le0\)

\(\Leftrightarrow0\le x\le4\)

b, \(\left|x-5\right|\le4\)

\(\Leftrightarrow\left(\left|x-5\right|\right)^2\le16\)

\(\Leftrightarrow x^2-10x+25\le16\)

\(\Leftrightarrow x^2-10x+9\le0\)

\(\Leftrightarrow1\le x\le9\)

\(\Rightarrow x\in\left\{1;2;3;4;5;6;7;8;9\right\}\)

c, \(\left|2x+1\right|< 3x\)

TH1: \(x\ge-\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow2x+1< 3x\)

\(\Leftrightarrow x>1\)

\(\Rightarrow\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

TH2: \(x< -\dfrac{1}{2}\)

\(\left|2x+1\right|< 3x\)

\(\Leftrightarrow-2x-1< 3x\)

\(\Leftrightarrow x>-\dfrac{1}{5}\left(l\right)\)

Vậy \(\left\{{}\begin{matrix}x\in Z\\x\in\left(1;2018\right)\end{matrix}\right.\)

d, \(\left|x+1\right|+\left|x\right|< 3\)

\(\Leftrightarrow x+1+x+2\left|x^2+x\right|< 9\)

\(\Leftrightarrow\left|x^2+x\right|< 4-x\)

Xét hai trường hợp để phá dấu giá trị tuyệt đối

e, Tương tự câu d

a, \(\left|3x+1\right|>2\)

\(\Leftrightarrow\left(\left|3x+1\right|\right)^2>4\)

\(\Leftrightarrow9x^2+6x+1>4\)

\(\Leftrightarrow9x^2+6x-3>0\)

\(\Leftrightarrow3\left(3x-1\right)\left(x+1\right)>0\)

\(\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{1}{3}\\x< -1\end{matrix}\right.\)

b, \(\left|2x-1\right|\le1\)

\(\Leftrightarrow\left(\left|2x-1\right|\right)^2\le1\)

\(\Leftrightarrow4x^2-4x+1\le1\)

\(\Leftrightarrow4x\left(x-1\right)\le0\)

\(\Leftrightarrow0\le x\le1\)

c, ĐK: \(x\ne13\)

\(\left|\dfrac{2}{x-13}\right|>\dfrac{8}{9}\)

\(\Leftrightarrow\dfrac{1}{\left|x-13\right|}>\dfrac{4}{9}\)

\(\Leftrightarrow4\left|x-13\right|< 9\)

\(\Leftrightarrow16\left(x^2-26x+169\right)< 81\)

\(\Leftrightarrow16x^2-416x+2623< 0\)

\(\Leftrightarrow\dfrac{43}{4}< x< \dfrac{61}{4}\)

\(\Rightarrow\) Có hai giả trị thỏa mãn yêu cầu bài toán

\(\left(3^x;3^y;3^z\right)=\left(a;b;c\right)\Rightarrow\left\{{}\begin{matrix}a;b;c>0\\ab+bc+ca=abc\end{matrix}\right.\)

BĐT cần chứng minh trở thành:

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

Thật vậy, ta có:

\(VT=\dfrac{a^3}{a^2+abc}+\dfrac{b^3}{b^2+abc}+\dfrac{c^3}{c^2+abc}\)

\(VT=\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{b^3}{\left(a+b\right)\left(b+c\right)}+\dfrac{c^3}{\left(a+c\right)\left(b+c\right)}\)

Áp dụng AM-GM:

\(\dfrac{a^3}{\left(a+b\right)\left(a+c\right)}+\dfrac{a+b}{8}+\dfrac{a+c}{8}\ge\dfrac{3a}{4}\)

Làm tương tự với 2 số hạng còn lại, cộng vế với vế rồi rút gọn, ta sẽ có đpcm