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AH
Akai Haruma
Giáo viên
28 tháng 6 2019
Lời giải:
a) ĐK: \(x>0; x\neq 25; x\neq 36\)
PT \(\Rightarrow (\sqrt{x}-2)(\sqrt{x}-6)=(\sqrt{x}-5)(\sqrt{x}-4)\)
\(\Leftrightarrow x-8\sqrt{x}+12=x-9\sqrt{x}+20\)
\(\Leftrightarrow \sqrt{x}=8\Rightarrow x=64\) (thỏa mãn)
Vậy.......
b)
ĐK: \(x\geq \frac{-1}{2}\)
PT \(\Leftrightarrow \sqrt{9(2x+1)}-\sqrt{4(2x+1)}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow 3\sqrt{2x+1}-2\sqrt{2x+1}+\frac{1}{3}\sqrt{2x+1}=4\)
\(\Leftrightarrow \frac{4}{3}\sqrt{2x+1}=4\Leftrightarrow \sqrt{2x+1}=3\)
\(\Rightarrow x=\frac{3^2-1}{2}=4\) (thỏa mãn)
c)
ĐK: \(x\geq 2\)
PT \(\Leftrightarrow \sqrt{4(x-2)}-\frac{1}{2}\sqrt{x-2}+\sqrt{9(x-2)}=9\)
\(\Leftrightarrow 2\sqrt{x-2}-\frac{1}{2}\sqrt{x-2}+3\sqrt{x-2}=9\)
\(\Leftrightarrow \frac{9}{2}\sqrt{x-2}=9\Leftrightarrow \sqrt{x-2}=2\Rightarrow x=2^2+2=6\) (thỏa mãn)
CB
3 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\Rightarrow a+b+c=18\)
Có: BDT
\(\Leftrightarrow\sum_{cyc}\left(\frac{b+c+5}{a+1}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{a+b+c-a+5}{a+1}\right)\ge\frac{51}{7}\)(1)
Đặt tiếp tục: \(\left\{{}\begin{matrix}m=a+1\\n=b+1\\p=c+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sum_{cyc}\left(\frac{24-m}{m}\right)\ge\frac{51}{7}\)
\(\Leftrightarrow\sum_{cyc}\left(\frac{24}{m}-1\right)\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge21\cdot\frac{3}{7}=9\)
\(\left(\frac{m}{n}-2+\frac{n}{m}\right)+\left(\frac{p}{m}-2+\frac{m}{p}\right)+\left(\frac{n}{p}-2+\frac{p}{n}\right)\ge0\)
\(\Leftrightarrow\frac{\left(m-n\right)^2}{mn}+\frac{\left(p-m\right)^2}{pm}+\frac{\left(n-p\right)^2}{pn}\ge0\)(đúng)
CB
4 tháng 6 2019
Đặt: \(\left\{{}\begin{matrix}x=a\\2y=b\\3z=c\end{matrix}\right.\)
BĐT
\(\Leftrightarrow\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{a+b+c-a+5}{a+1}+\frac{a+c+b-b+5}{b+1}+\frac{a+b+c-c+5}{c+1}\ge\frac{51}{7}\)
\(\Leftrightarrow\frac{24-\left(a+1\right)}{a+1}+\frac{24-\left(b+1\right)}{b+1}+\frac{24-\left(c+1\right)}{c+1}\ge\frac{51}{7}\)(1)
Đặt tiếp: \(\left\{{}\begin{matrix}a+1=m\\b+1=n\\c+1=p\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=m-1\\b=n-1\\c=p-1\end{matrix}\right.\)
(1)\(\Leftrightarrow\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{51}{7}\)
\(\Leftrightarrow24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{72}{7}\)
\(\Leftrightarrow\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\ge\frac{3}{7}\)
\(\Leftrightarrow\left(m+n+p\right)\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)\ge\frac{3}{7}\left(m+n+p\right)\)( do m+n+p>0)
\(\Leftrightarrow3+\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{m}{p}+\frac{p}{m}\ge\frac{3}{7}\left[\left(a+b+c\right)+3\right]\)
\(\Leftrightarrow\frac{m}{n}+\frac{n}{m}+\frac{p}{n}+\frac{n}{p}+\frac{p}{m}+\frac{m}{p}-6\ge0\)
Tới đây chắc bn làm đc rồi
18 tháng 9 2016
Đặt \(\hept{\begin{cases}a=x\\b=2y\\c=3z\end{cases}}\) => a + b + c = 18
\(P=\frac{2y+3z+5}{1+x}+\frac{3z+x+5}{1+2y}+\frac{x+2y+5}{1+3z}=\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+b+5}{c+1}\)
Lại đặt \(\hept{\begin{cases}m=a+1\\n=b+1\\p=c+1\end{cases}}\Rightarrow\hept{\begin{cases}a=m-1\\b=n-1\\c=p-1\end{cases}}\)
Ta có : \(\frac{b+c+5}{a+1}+\frac{a+c+5}{b+1}+\frac{a+c+5}{c+1}=\frac{24-m}{m}+\frac{24-n}{n}+\frac{24-p}{p}\)
\(=24\left(\frac{1}{m}+\frac{1}{n}+\frac{1}{p}\right)-3\ge\frac{24.9}{m+n+p}-3=\frac{24.9}{\left(a+1\right)+\left(b+1\right)+\left(b+1\right)}-3\)
\(=\frac{24.9}{18+3}-3=\frac{51}{7}\)
21 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
PN
1
AH
Akai Haruma
Giáo viên
12 tháng 8 2020
Lời giải:
ĐKXĐ: $x\geq -2$
PT $\Leftrightarrow \frac{1}{5}\sqrt{25(x+2)}-5\sqrt{x+2}+\sqrt{9(x+2)}+9=0$
$\Leftrightarrow \sqrt{x+2}-5\sqrt{x+2}+3\sqrt{x+2}+9=0$
$\Leftrightarrow \sqrt{x+2}=9$
$\Rightarrow x=9^2-2=79$ (thỏa mãn)
Vậy.........
\(\frac{x^9-1}{x^9+1}=1+\frac{-2}{x^9+1}\)
Mà \(\frac{x^9-1}{x^9+1}=7\)
Nên:\(1+\frac{-2}{x^9+1}=7\Leftrightarrow x^9+1=-\frac{1}{3}\Leftrightarrow x^9=-\frac{4}{3}\)
Suy ra: \(\frac{x^{18}-1}{x^{18}+1}=\frac{\left(x^9\right)^2-1}{\left(x^9\right)^2+1}=\text{tự thế vào}\)
Toán lớp 9 à ? Mk hết hè này mới nên lớp 6 nên chưa học , Chúc bn nhanh tìm đc đáp án đúq
k mk nhék ^.^