phân tích thành nhân tử
\(12-\sqrt{x}-x\)
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\(\frac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}-\frac{1-a}{1+\sqrt{a}}\)
ĐKXĐ : \(a\ge0\)
\(=\frac{\left(2+\sqrt{a}-\sqrt{a}+1\right)\left(2+\sqrt{a}+\sqrt{a}+1\right)}{2\sqrt{a}+3}-\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1+\sqrt{a}}\)
\(=\frac{3\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}-\left(1-\sqrt{a}\right)\)
\(=3-1+\sqrt{a}\)
\(=\sqrt{a}+2\)
Còn như nào bạn tự giải nốt nhé *)
Ta có : \(\frac{1}{P}=\frac{a}{\sqrt{a-2}}=\frac{a-2+2}{\sqrt{a-2}}=\sqrt{a-2}+\frac{2}{\sqrt{a-2}}\)
Áp dụng BĐT Cauchy :
\(\sqrt{a-2}+\frac{2}{\sqrt{a-2}}\ge2\sqrt{\sqrt{a-2}.\frac{2}{\sqrt{a-2}}}=2\sqrt{2}\)
\(\Leftrightarrow P\ge\frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{a-2}=\frac{2}{\sqrt{a-2}}\Leftrightarrow a-2=2\Leftrightarrow a=4\)
Vậy maxP =\(\frac{\sqrt{2}}{4}\)<=> a = 4
a) \(\sqrt{\left(2x-1\right)^2}=3\)
⇔ \(\left|2x-1\right|=3\)
⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
ĐKXĐ : \(x\ge0\)
⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)
⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)
⇔ \(7\sqrt{x}-6\sqrt{x}=5\)
⇔ \(\sqrt{x}=5\)
⇔ \(x=25\)( tm )
c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)
ĐKXĐ : \(x\ge-5\)
⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)
⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)
⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)
⇔ \(\frac{5}{4}\sqrt{x+5}=6\)
⇔ \(\sqrt{x+5}=\frac{24}{5}\)
⇔ \(x+5=\frac{576}{25}\)
⇔ \(x=\frac{451}{25}\left(tm\right)\)
Đặt \(A=\frac{x}{y^4+2}+\frac{y}{z^42}+\frac{z}{x^4+2}\ge1\)
\(A=\frac{y^4}{x+2}+\frac{z^4}{y+2}+\frac{x^4}{z+2}\ge1\)
Còn lại thì bạn tính tổng nha! Lớn hơn hoặc bằng 1 là được :))
a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)
ĐK : x ≥ 0
⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)
⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)
⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)
⇔ \(-3\sqrt{x}=-5\)
⇔ \(\sqrt{x}=15\)
⇔ \(x=225\)( tm )
b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)
ĐK : x ≤ 3
⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)
⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)
⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)
⇔ \(3\sqrt{3-x}=6\)
⇔ \(\sqrt{3-x}=2\)
⇔ \(3-x=4\)
⇔ \(x=-1\)( tm )
c) \(\sqrt{9x^2+12x+4}=4\)
⇔ \(\sqrt{\left(3x+2\right)^2}=4\)
⇔ \(\left|3x+2\right|=4\)
⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)
d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)
ĐK : x ≥ 1
⇔ \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)
⇔ \(\sqrt{x-1}=1\)
⇔ \(x-1=1\)
⇔ \(x=2\)( tm )
\(12-\sqrt{x}-x\)
ĐK : x ≥ 0
\(=12-4\sqrt{x}+3\sqrt{x}-x\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(3-\sqrt{x}\right)\left(4+\sqrt{x}\right)\)
\(12-\sqrt{x}-x=\left(12-4\sqrt{x}\right)+\left(3\sqrt{x}-x\right)\)
\(=4\left(3-\sqrt{x}\right)+\sqrt{x}\left(3-\sqrt{x}\right)\)
\(=\left(4+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)