chj thông cảm cho em nha

em lp 6

chúc chj HT

ĐKXĐ: x≥0;x≠1x≥0;x≠1

P=15√x−11(√x−1)(√x+3)−(3√x−2)(√x+3)(√x−1)(√x+3)−(2√x+3)(√x−1)(√x−1)(√x+3)P=15x−11(x−1)(x+3)−(3x−2)(x+3)(x−1)(x+3)−(2x+3)(x−1)(x−1)(x+3)

=15√x−11−3x−7√x+6−2x−√x+3(√x−1)(√x+3)=15x−11−3x−7x+6−2x−x+3(x−1)(x+3)

=−5x+7√x−2(√x−1)(√x+3)=−(√x−1)(5√x−2)(√x−1)(√x+3)=2−5√x√x+3=−5x+7x−2(x−1)(x+3)=−(x−1)(5x−2)(x−1)(x+3)=2−5xx+3

P=12⇒2−5√x√x+3=12⇒4−10√x=√x+3P=12⇒2−5xx+3=12⇒4−10x=x+3

⇒11√x=1⇒√x=111⇒x=1121⇒11x=1⇒x=111⇒x=1121

P=17−5(√x+3)√x+3=−5+17√x+3P=17−5(x+3)x+3=−5+17x+3

Do √x+3≥3⇒−5+17√x+3≤−5+173=23x+3≥3⇒−5+17x+3≤−5+173=23

Pmax=23Pmax=23 khi x=0x=0, hình như bạn nhầm đề, ko có GTNN đâu, chỉ có GTLN thôi

P=−5+17√x+3⇒P=−5+17x+3⇒ để P nguyên thì √x+3=Ư(17)x+3=Ư(17)

Mà √x+3≥3⇒√x+3=17x+3≥3⇒x+3=17

⇒√x=14⇒x=196

Ap dụng bất đẳng thức BDT Caucchy Schwarz ta có :

\(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2zx}+\frac{z^2}{z^2+2xy}\)

\(=\frac{\left(x+y+z\right)^2}{x^2+2yz+y^2+2zx+z^2+2xy}\)

\(=\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)

Answer:

\(\left(x-5\right)^2+\left(x+4\right)\left(3-x\right)=4\)

\(\Rightarrow x^2-10x+25+3x-x^2+12-4x-4=0\)

\(\Rightarrow-11x+33=0\)

\(\Rightarrow x=3\)

\(a,=\frac{x^3+26x-19}{\left(x-3\right)\left(x+1\right)}+\frac{2x}{x+1}-\frac{x+3}{x-3}\)

\(=\frac{x^3+26x-19+2x^2-6x-x^2-4x-4}{\left(x-3\right)\left(x+1\right)}\)

\(=\frac{x^3+x^2+16x-23}{\left(x-3\right)\left(x+1\right)}\)