Do \(\hept{\begin{cases}\sin B< 1\\\cos B< 1\end{cases}}\) nên \(\hept{\begin{cases}\sin^{2018}B< \sin^{2017}B< ...< \sin^2B\\\cos^{2018}B< \cos^{2017}B< ...< \cos^2B\end{cases}}\)

\(\Rightarrow\)\(\sin^{2018}B+\cos^{2018}B< \sin^2B+\cos^2B=1\)

hello

\(a,x=7-4\sqrt{3}=4-2.2\sqrt{3}+3\) (Thỏa mãn ĐKXĐ)

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

\(B=\frac{2}{\sqrt{x}-2}=\frac{2}{\sqrt{\left(2-\sqrt{3}\right)^2}-2}\)

\(=\frac{2}{2-\sqrt{3}-2}=-\frac{2\sqrt{3}}{3}\)

\(b,P=\frac{B}{A}=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}\right)\)

\(=\frac{2}{\sqrt{x}-2}:\left(\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\)

\(=\frac{2}{\sqrt{x}-2}:\frac{\sqrt{x}+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{2}{\sqrt{x}-2}:\frac{2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{2}{\sqrt{x}-2}.\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{2\left(\sqrt{x}+1\right)}\)

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

\(P=\frac{4}{3}\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}+1}=\frac{4}{3}\)

\(\Leftrightarrow3\left(\sqrt{x}+2\right)=4\left(\sqrt{x}+1\right)\)

\(\Leftrightarrow3\sqrt{x}+6=4\sqrt{x}+4\)

\(\Leftrightarrow6-4=4\sqrt{x}-3\sqrt{x}\)

\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)(ko thỏa mãn ĐKXĐ)

=>pt vo nghiệm

d,\(\left(\sqrt{x}+1\right)P-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)

\(\Leftrightarrow\left(\sqrt{x}+1\right)\frac{\sqrt{x}+2}{\sqrt{x}+1}-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)

\(\Leftrightarrow\sqrt{x}+2-\sqrt{x}-4\sqrt{x-1}+26=-6x+10\sqrt{5x}\)

\(\Leftrightarrow-4\sqrt{x-1}+28=-6x+10\sqrt{5x}\)

\(\Leftrightarrow x=5\)

\(\hept{\begin{cases}x^2+y^2=5\\x^3+2y^2=10x-10y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\sqrt{5-y^2}\\\left(\sqrt{5-y^2}\right)^3+2y^2=10\sqrt{5-y^2}-10y\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\sqrt{5-y^2}\\y=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\sqrt{5-1^2}=\sqrt{4}=2\\y=1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=2\\y=1\end{cases}}\)

sai rồi bạn ạ còn một GT nữa cơ bạn thử xét x=-2 y=-1 xem

a) \(\left|2x-3\right|=\left|1-x\right|\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3=1-x\\2x-3=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)

b) \(x^2-4x\le5\)

\(\Leftrightarrow x^2-4x-5\le0\)

\(\Leftrightarrow x^2-5x+x-5\le0\)

\(\Leftrightarrow x\left(x-5\right)+\left(x-5\right)\le0\)

\(\Leftrightarrow\left(x+1\right)\left(x-5\right)\le0\)

Đến đây dễ r

c) \(2x\left(2x-1\right)\le2x-1\)

\(\Leftrightarrow2x\left(2x-1\right)-\left(2x-1\right)\le0\)

\(\Leftrightarrow\left(2x-1\right)^2\le0\)

Mà \(\left(2x-1\right)^2\ge0\)nên 2x - 1=0

\(VT=\frac{a^2}{ab^2+abc+ac^2}+\frac{b^2}{c^2b+abc+a^2b}+\frac{c^2}{a^2c+abc+b^2c}\)

Áp dụng BĐT Cauchy dạng phân thức 

\(\Rightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b\right)+abc+ac\left(a+c\right)+abc+bc\left(b+c\right)+abc}\)

\(\Leftrightarrow VT\ge\frac{\left(a+b+c\right)^2}{ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)}\)

\(=\frac{\left(a+b+c\right)^2}{\left(a+b+c\right)\left(ab+bc+ac\right)}\)

\(\Leftrightarrow VT\ge\frac{a+b+c}{ab+bc+ac}\left(đpcm\right)\)

Dấu "=" xảy ra khi a=b=c

Chúc bạn học tốt !!!