Vẽ tam giác ABC vuông tại A có \(\dfrac{AB}{BC}=\dfrac{2}{7}\), tính \(\widehat{B};\widehat{C}=?\)

Áp dụng hệ thức lượng vào tam giác vuông có:

\(sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{2}{7}\Leftrightarrow\widehat{C}\approx16,6^0\)

\(cos\widehat{B}=\dfrac{AB}{BC}=\dfrac{2}{7}\Leftrightarrow\widehat{B}\approx73,4^0\)

Vậy...

𝔊𝔦𝔳𝔢 𝔭𝔢𝔞𝔠𝔢 𝔞 𝔠𝔥𝔞𝔫𝔠𝔢
𝔏𝔢𝔱 𝔱𝔥𝔢 𝔣𝔢𝔞𝔯 𝔶𝔬𝔲 𝔥𝔞𝔳𝔢 𝔣𝔞𝔩𝔩 𝔞𝔴𝔞𝔶

\(\sqrt{144a}+3\sqrt{9a^2}+2a=\sqrt{12^2.a}+3\sqrt{\left(3a\right)^2}+2a\)

\(=12\sqrt{a}+3.\left|3a\right|+2a\)

\(=12\sqrt{a}+3.3a+2a\) ( do \(a\ge0\) )

\(=12\sqrt{a}+11a\)

\(\sqrt{\dfrac{1}{9}a}-\sqrt{\dfrac{2}{50}x}+3\sqrt{\dfrac{4}{16}x}\)

\(=\sqrt{\left(\dfrac{1}{3}\right)^2x}-\sqrt{\left(\dfrac{1}{5}\right)^2x}+3\sqrt{\left(\dfrac{2}{4}\right)^2x}\)

\(=\dfrac{1}{3}\sqrt{x}-\dfrac{1}{5}\sqrt{x}+3.\dfrac{2}{4}\sqrt{x}=\dfrac{49}{30}\sqrt{x}\)

2) ĐK: \(x\ge0;x\ne4\)

\(A=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-9-\left(x-4\right)+2\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{2}{\sqrt{x}-2}\)

Vậy...

1) \(\sqrt{14+6\sqrt{5}}=\sqrt{5+2.3.\sqrt{5}+9}=\sqrt{\left(\sqrt{5}+3\right)^2}=\sqrt{5}+3\)

Thay \(x=14+6\sqrt{5}\) vào B :

\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{\sqrt{14+6\sqrt{5}}+2}{\sqrt{14+6\sqrt{5}}-2}=\dfrac{3+\sqrt{5}+2}{3+\sqrt{5}-2}\)\(=\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\)

\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)\(=\dfrac{5\sqrt{5}-5+5-\sqrt{5}}{4}=\sqrt{5}\)

Vậy...

3) \(B=-\dfrac{5}{3}\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-2}=-\dfrac{5}{3}\left(ĐK:x\ge0;x\ne4\right)\)

\(\Rightarrow3\sqrt{x}+6=-5\sqrt{x}+10\) \(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{1}{4}\) (tm)

Vậy...

4) \(A\le-\dfrac{2}{3}\Leftrightarrow\dfrac{2}{\sqrt{x}-2}\le-\dfrac{2}{3}\) \(\Leftrightarrow\dfrac{2}{2-\sqrt{x}}\ge\dfrac{2}{3}\Leftrightarrow2-\sqrt{x}\le3\)

\(\Leftrightarrow\sqrt{x}\ge-1\left(lđ\right)\)

Vậy \(\forall x\) thì \(A\le-\dfrac{2}{3}\)

5) \(A>B\Leftrightarrow\dfrac{2}{\sqrt{x}-2}>\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)

TH1: \(\sqrt{x}-2>0\Leftrightarrow x>4\)

BPT \(\Leftrightarrow2>\sqrt{x}+2\Leftrightarrow0>\sqrt{x}\left(vn\right)\) 

TH2: \(\sqrt{x}-2< 0\Leftrightarrow x< 4\)

BPT \(\Leftrightarrow2< \sqrt{x}+2\Leftrightarrow\sqrt{x}>0\Leftrightarrow x>0\)

Vậy \(A< B\Leftrightarrow0< x< 4\Rightarrow x=\left\{1;2;3\right\}\)

6) Đk của B là \(x\ge0;x\ne4\)

\(\dfrac{3}{B}=\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}+2}=\dfrac{3\left(\sqrt{x}+2\right)-12}{\sqrt{x}+2}=3-\dfrac{12}{\sqrt{x}+2}\)

\(\dfrac{3}{B}\in Z\Leftrightarrow\dfrac{12}{\sqrt{x}+2}\in Z\). Ta có \(\sqrt{x}+2\ge2\Leftrightarrow0< \dfrac{12}{\sqrt{x}+2}\le6\)

TH1: \(\dfrac{12}{\sqrt{x}+2}=1\) \(\Leftrightarrow x=100\left(tmđk\right)\)

Tương tự ta được các x thỏa mãn là 16; 1; 0,16; 0

Vậy \(x\in\left\{0,\dfrac{4}{25},1,16,100\right\}\)

[бяїgнтёя тнап а бгцё $кч]

Đk: \(x\ne\dfrac{\pi}{2}+k\pi\left(k\in Z\right)\)

PT \(\Leftrightarrow2\left(sinx.\dfrac{\sqrt{2}}{2}-cosx.\dfrac{\sqrt{2}}{2}\right)^2=2sin^2x-\dfrac{sinx}{cosx}\)

\(\Leftrightarrow\left(sinx-cosx\right)^2=2sin^2x-\dfrac{sinx}{cosx}\)

\(\Leftrightarrow1-2.sinx.cosx=2sin^2x-\dfrac{sinx}{cosx}\)

\(\Leftrightarrow cosx-2sinx.cos^2x=2sin^2x.cosx-sinx\)

\(\Leftrightarrow\left(cosx+sinx\right)-2sinx.cosx\left(cosx+sinx\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\1-2sinx.cosx=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\sin2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\) ( k nguyên ) (tmđk)

Vậy...

𝐼 𝒹𝑜𝓃'𝓉 𝒸𝒶𝓇𝑒 𝒽𝑜𝓌 𝓁𝑜𝓃𝑔 𝒾𝓉 𝓉𝒶𝓀𝑒𝓈

1) Đk: \(x\ge0;x\ne1\)

\(A=4\Leftrightarrow\dfrac{x-4}{\sqrt{x}-1}=4\Leftrightarrow\dfrac{x-4}{\sqrt{x}-1}-4=0\)

\(\Rightarrow x-4-4\left(\sqrt{x}-1\right)=0\Leftrightarrow x-4\sqrt{x}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\) (tm)

Vậy...

2) ĐK: \(x\ge0;x\ne4\)

\(B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}+1}{3}\)

\(=\dfrac{x-1-\left(x-4\right)}{3\left(\sqrt{x}-2\right)}=\dfrac{1}{\sqrt{x}-2}\)

Vậy...

3) \(P=\dfrac{1}{AB}=\dfrac{1}{\dfrac{x-4}{\sqrt{x}-1}.\dfrac{1}{\sqrt{x}-2}}\)\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)\(=1-\dfrac{3}{\sqrt{x}+2}\)

Có \(\sqrt{x}+2\ge2\Leftrightarrow\dfrac{3}{\sqrt{x}+2}\le\dfrac{3}{2}\) \(\Leftrightarrow1-\dfrac{3}{\sqrt{x}+2}\ge1-\dfrac{3}{2}\Leftrightarrow P\ge-\dfrac{1}{2}\)

Suy ra \(P_{min}=-\dfrac{1}{2}\Leftrightarrow x=0\)

Vậy...

4) \(P-\dfrac{3}{2}=1-\dfrac{3}{\sqrt{x}+2}-\dfrac{3}{2}=-\dfrac{1}{2}-\dfrac{3}{\sqrt{x}+2}< 0\)

Suy ra \(P< \dfrac{3}{2}\)

𝕴 𝖐𝖓𝖔𝖜 𝖎𝖘 𝖜𝖊'𝖗𝖊 𝖌𝖔𝖎𝖓𝖌 𝖍𝖔𝖒𝖊

\(XT^2=TY.TZ\Leftrightarrow2XT^2+TY^2+TZ^2=TY^2+TZ^2+2TY.TZ\)

\(\Leftrightarrow\left(XT^2+TY^2\right)+\left(XT^2+TZ^2\right)=\left(TY+TZ\right)^2\)

\(\Leftrightarrow XZ^2+XY^2=YZ^2\)

Suy ra tam giác XYZ vuông tại X.

ღღღ ꋖꌅờꂑ ꀯꁲꂦ ꒒ắꂵ ꋰóꋊꁅ ꂵâꐞ ღღღ

\(\overrightarrow{DM}.\overrightarrow{A'N}=\left(\overrightarrow{DA}+\overrightarrow{AM}\right)\left(\overrightarrow{A'B'}+\overrightarrow{B'N}\right)\)

\(=\overrightarrow{DA}.\overrightarrow{A'B'}+\overrightarrow{AM}.\overrightarrow{A'B'}+\overrightarrow{DA}.\overrightarrow{B'N}+\overrightarrow{AM}.\overrightarrow{B'N}\)

( chứng minh được \(DA\perp A'B',AM\perp B'N\) )

\(=0+\dfrac{1}{2}\overrightarrow{AB}.\overrightarrow{AB}+\overrightarrow{C'B'}.\left(-\dfrac{1}{2}\overrightarrow{C'B'}\right)+0\)

\(=\dfrac{1}{2}AB^2-\dfrac{1}{2}C'B'^2=0\)

Suy ra \(DM\perp A'N\)

Ý A

a) \(4\times\dfrac{1}{4^2}+25\times\left[\dfrac{3^3}{4^3}:\dfrac{5^3}{4^3}\right]:\dfrac{3^3}{2^3}\)

\(=\dfrac{1}{4}+5^2\times\dfrac{3^3}{4^3}\times\dfrac{4^3}{5^3}\times\dfrac{2^3}{3^3}\)

\(=\dfrac{1}{4}+\dfrac{2^3}{5}=\dfrac{1}{4}+\dfrac{8}{5}=\dfrac{37}{20}\)

b) \(2^3+3\times\left(\dfrac{1}{2}\right)^{0-1}+\left[\left(-2\right)^2:\dfrac{1}{2}\right]-8\)

\(=8+3\times\left(2^{-1}\right)^{-1}+2^2\times2-8\)

\(=3\times2+2^3=14\)

Muộn rùi ngủ thôi không mai lớn có một khúc à :v 

Để B tồn tại \(\Leftrightarrow2m< 3m+1\Leftrightarrow m>-1\)

TH1: \(10\le3m+1\) \(\Leftrightarrow m\ge3\)

\(A\cap B=[2m;10)\) có đúng ba số nguyên khi \(6< 2m\le7\) \(\Leftrightarrow3< m\le\dfrac{7}{2}\) ( tm đk )

TH2: \(3m+1< 10\) \(\Leftrightarrow m< 3\)

\(A\cap B=\left[2m;3m+1\right]\) có đúng ba số nguyên khi 

Trường hợp m nguyên thì \(2m+2=3m+1\Leftrightarrow m=1\) (thỏa mãn)

Trường hợp m là số thực thì rộng lắm...

PT \(\Leftrightarrow-2\left(1-2.sin^2\dfrac{x}{2}\right)-\sqrt{3}.cos2x=-1+2\left(cosx.cos\dfrac{3\pi}{4}-sinx.sin\dfrac{3\pi}{4}\right)^2\)

\(\Leftrightarrow-2.cosx-\sqrt{3}.cos2x=-1+2\left(cosx.-\dfrac{\sqrt{2}}{2}-sinx.\dfrac{\sqrt{2}}{2}\right)^2\)

\(\Leftrightarrow-2cosx-\sqrt{3}.cos2x=-1+\left(sinx+cosx\right)^2\)

\(\Leftrightarrow-2cosx=2sinx.cosx+\sqrt{3}cos2x\)

\(\Leftrightarrow-2cosx=sin2x+\sqrt{3}cos2x\)

\(\Leftrightarrow cos\left(\pi-x\right)=\dfrac{1}{2}.sin2x+\dfrac{\sqrt{3}}{2}.cos2x\)

\(\Leftrightarrow cos\left(\pi-x\right)=sin\left(2x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow cos\left(\pi-x\right)=cos\left(\dfrac{\pi}{6}-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\pi-x=\dfrac{\pi}{6}-2x+k2\pi\\\pi-x=-\dfrac{\pi}{6}+2x+k2\pi\end{matrix}\right.\) ( k nguyên )

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5\pi}{6}+k2\pi\\x=\dfrac{7\pi}{18}-\dfrac{k2\pi}{3}\end{matrix}\right.\) ( k nguyên )

Vậy...