ĐKXĐ: \(-2\le x\le2\)

Với \(-2\le x\le\frac{2}{3}\Rightarrow6x-4\le0\Rightarrow VT\ge VP\) BPT luôn đúng

- Với \(\frac{2}{3}\le x\le3\) ta có:

\(VT^2=\left(\sqrt{2x+4}+2\sqrt{2-x}\right)^2=12-2x+4\sqrt{2\left(4-x^2\right)}\ge8\)

\(\Rightarrow VT\ge2\sqrt{2}\)

\(VP=\frac{6x-4}{5\sqrt{x^2+1}}< \frac{6x-4}{5}\le\frac{12-4}{5}=\frac{8}{5}< 2\sqrt{2}\)

\(\Rightarrow VT>VP\)

Vậy BPT luôn đúng với mọi \(x\in\left[-2;2\right]\Rightarrow\left\{{}\begin{matrix}a=-2\\b=2\end{matrix}\right.\) \(\Rightarrow P=-10\)

\(-x^2-2\left(m-1\right)x+2m-1>0\)

\(\Leftrightarrow x^2+2\left(m-1\right)x-2m+1< 0\)

\(f\left(x\right)=x^2+2\left(m-1\right)x-2m+1\)

Yêu cầu bài toán thỏa mãn khi \(f\left(x\right)=0\) có hai nghiệm phân biệt thỏa mãn \(x_1\le0< 1\le x_2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'=\left(m-1\right)^2+2m-1>0\\f\left(1\right)\le0\\f\left(0\right)\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2>0\\1+2\left(m-1\right)-2m+1\le0\\-2m+1\le0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m\ne0\\m\ge\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow m\ge\dfrac{1}{2}\)

\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)

\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)

\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)

\(\Leftrightarrow2x-13< 0\)

\(\Leftrightarrow x< \dfrac{13}{2}\)

KL...............

\(b.5+\dfrac{x+4}{5}< x-\dfrac{x-2}{2}+\dfrac{x+3}{3}\)

\(\Leftrightarrow\dfrac{150+6\left(x+4\right)}{30}< \dfrac{30x-15\left(x-2\right)+10\left(x+3\right)}{30}\)

\(\Leftrightarrow150+6x+24< 30x-15x+30+10x+30\)

\(\Leftrightarrow6x-30x+15x-10x+150+24-30-30< 0\)

\(\Leftrightarrow-19x+114< 0\)

\(\Leftrightarrow x>6\)

KL..................

Câu 4 :

Ta có :

\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)

\(=\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\)

Theo BĐT Bu - nhi a - cốp xki ta có :

\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

\(\Leftrightarrow\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\ge\left(\sqrt{\dfrac{3\left(1-x\right)}{1-x}}+\sqrt{\dfrac{4x}{x}}\right)^2=\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)

Dấu \("="\) xảy ra khi \(\dfrac{3}{\left(1-x\right)^2}=\dfrac{4}{x^2}\)

\(\Leftrightarrow3x^2=4x^2-8x+4\)

\(\Leftrightarrow x^2-8x+4=0\)

\(\Delta=64-16=48>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)

Vậy GTNN của\(A=7+4\sqrt{3}\) khi \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)

câu nào các bn lm đc thid cmt giúp mik vs ak.1 câu thôi cug đc

lop 7 lam gi co nghiem voi da thuc ha ban

Đề thi HSG lớp 7 đó bạn

Câu 1:

\(\Leftrightarrow\left\{{}\begin{matrix}13x>\dfrac{7}{3}\\4x-16< 3x-14\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\dfrac{7}{39}\\x< 2\end{matrix}\right.\Leftrightarrow\dfrac{7}{39}< x< 2\)

mà x nguyên

nên x=1

Câu 2: 

\(\Leftrightarrow\left\{{}\begin{matrix}2x< 4\\mx>2-m\end{matrix}\right.\)

=>x<2 và mx>2-m

Nếu m=0 thì bất phươg trình vô nghiệm

Nếu m<>0 thì BPT sẽ tương đương với:

\(\left\{{}\begin{matrix}x< 2\\x>\dfrac{2-m}{m}\end{matrix}\right.\)

Để BPT vô nghiệm thì 2-m/m>=2

=>\(\dfrac{2-m}{m}-2>=0\)

=>\(\dfrac{2-m-2m}{m}>=0\)

=>\(\dfrac{3m-2}{m}< =0\)

=>0<m<=2/3

ĐK \(\hept{\begin{cases}x\ge5\\x\le-2\end{cases}}\)

\(\Rightarrow x^2-3x-10< x^2-4x+4\)

<=> x<14

=> (a;b)=(5;14)

=> a+b=19