1) \(\sqrt[]{3x}-5\sqrt[]{12x}+7\sqrt[]{27x}=12\)

\(\Leftrightarrow\sqrt[]{3x}\left(1-5.2+7.3\right)=12\)

\(\Leftrightarrow\sqrt[]{3x}.12=12\)

\(\Leftrightarrow\sqrt[]{3x}=1\left(x\ge0\right)\)

\(\Leftrightarrow3x=1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

2) \(\sqrt[3]{x^2+2}=3\)

\(\Leftrightarrow\left(\sqrt[3]{x^2+2}\right)^3=3^3\)

\(\Leftrightarrow x^2+2=27\)

\(\Leftrightarrow x^2=25\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\left(d\right):y=ax+b\\\left(d_1\right):y=3x+2\end{matrix}\right.\)

\(\left(d\right)//\left(d_1\right)\Leftrightarrow\left\{{}\begin{matrix}a=3\\b\ne2\end{matrix}\right.\) 

\(\Rightarrow\left(d\right):y=3x+b\)

\(\left(d\right)\cap Oy=A\left(0;-5\right)\)

\(\Leftrightarrow3.0+b=-5\)

\(\Leftrightarrow b=-5\)

Vậy \(\left(d\right):y=3x-5\)

1: Khi x=1,44 thì \(A=\dfrac{1.44+7}{1.2}=\dfrac{8.44}{1.2}=\dfrac{211}{30}\)

2: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)

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

\(=\dfrac{x-3\sqrt{x}+2x+5\sqrt{x}-3-2x+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)

3: \(S=\dfrac{1}{B}+A=\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{x+7}{\sqrt{x}}=\dfrac{x+\sqrt{x}+4}{\sqrt{x}}\)

\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}}\)

\(\Leftrightarrow S>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{4}{\sqrt{x}}}+1=4+1=5\)

Dấu = xảy ra khi \(\left(\sqrt{x}\right)^2=4\)

=>x=4

1:

Xét ΔABC vuông tại A có AH là đường cao

nên \(AB^2=BH\cdot BC\)

=>\(AB^2=2\cdot8=16\)

=>AB=4cm

BH+HC=BC

=>HC+2=8

=>HC=6(cm)

ΔAHB vuông tại H

=>\(HA^2+HB^2=AB^2\)

=>\(HA^2+2^2=4^2\)

=>HA^2=4^2-2^2=12

=>\(HA=2\sqrt{3}\left(cm\right)\)

ΔAHC vuông tại H

=>\(HA^2+HC^2=AC^2\)

=>\(AC^2=12+36=48\)

=>\(AC=4\sqrt{3}\left(cm\right)\)

2: ΔABK vuông tại A có AD là đường cao

nên \(BD\cdot BK=BA^2\left(1\right)\)

ΔABC vuông tại A có AH là đường cao

nên \(BH\cdot BC=BA^2\left(2\right)\)

Từ (1) và (2) suy ra \(BH\cdot BC=BD\cdot BK\)

3: BH*BC=BD*BK

=>BH/BK=BD/BC

Xét ΔBHD và ΔBKC có

\(\dfrac{BH}{BK}=\dfrac{BD}{BC}\)

\(\widehat{HBD}\) chung

Do đó: ΔBHD\(\sim\)ΔBKC

=>\(\dfrac{S_{BHD}}{S_{BKC}}=\left(\dfrac{BH}{BK}\right)^2\)

=>\(\dfrac{S_{BHD}}{S_{BKC}}=\dfrac{BH}{BK}\cdot\dfrac{BD}{BC}\)

\(=\dfrac{BH}{BC}\cdot\dfrac{BD}{BK}\)

\(=\dfrac{1}{4}\cdot\dfrac{BD}{BK}\)

\(cos^2ABK=cosABK\cdot cosABK=\dfrac{AB}{BK}\cdot\dfrac{BD}{BA}=\dfrac{BD}{BK}\)

=>\(\dfrac{S_{BHD}}{S_{BKC}}=\dfrac{1}{4}\cdot cos^2ABK\)

=>\(S_{BHD}=\dfrac{1}{4}\cdot S_{BKC}\cdot cos^2ABK\)

Sửa đề; NP=10cm

ΔMNP vuông tại M

=>\(MN^2+MP^2=NP^2\)

=>\(MP^2=10^2-6^2=64\)

=>MP=8(cm)

Xét ΔMNP vuông tại M có MH là đường cao

nên \(MH\cdot NP=MN\cdot MP\)

=>MH*10=6*8=48

=>MH=4,8(cm)

Xét ΔMNP vuông tại M có MH là đường cao

nên \(\left\{{}\begin{matrix}MN^2=NH\cdot NP\\PM^2=PH\cdot PN\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}NH=\dfrac{6^2}{10}=3,6\left(cm\right)\\PH=\dfrac{8^2}{10}=6,4\left(cm\right)\end{matrix}\right.\)

a) \(A=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x-3}{2x-x^2}\) (ĐK: \(x\ne0;x\ne\pm2;x\ne3\))

\(A=\left[-\dfrac{x+2}{x-2}+\dfrac{x-2}{x+2}-\dfrac{4x^2}{\left(x+2\right)\left(x-2\right)}\right]:\dfrac{x-3}{2x-x^2}\)

\(A=\left[\dfrac{\left(x-2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}-\dfrac{4x^2}{\left(x+2\right)\left(x-2\right)}\right]\cdot\dfrac{2x-x^2}{x-3}\)

\(A=\dfrac{x^2-4x+4-x^2-4x-4-4x^2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{2x-x^2}{x-3}\)

\(A=\dfrac{-4x^2-8x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}\)

\(A=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)

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

\(A=\dfrac{4x^2}{x-3}\)

b) Ta có: 

\(\left|x\right|=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

TH1: x=1

\(A=\dfrac{4\cdot1^2}{1-3}=\dfrac{4}{-2}=-2\)

TH2: x=-1

\(A=\dfrac{4\cdot\left(-1\right)^2}{-1-3}=\dfrac{4}{-4}=-1\)

a) \(A=\left(\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}-\dfrac{4x^2}{x^2-4}\right):\dfrac{x-3}{2x-x^2}\left(x\ne0;\pm2\right)\)

\(\Leftrightarrow A=\left[\dfrac{\left(2+x\right)^2-\left(2-x\right)^2+4x^2}{\left(2-x\right)\left(2+x\right)}\right]-:\dfrac{x-3}{x\left(2-x\right)}\)

\(\Leftrightarrow A=\left[\dfrac{4+4x+x^2-\left(4-4x+x^2\right)+4x^2}{\left(2-x\right)\left(2+x\right)}\right]-:\dfrac{x-3}{x\left(2-x\right)}\)

\(\Leftrightarrow A=\left[\dfrac{4x^2+8x}{\left(2-x\right)\left(2+x\right)}\right]:\dfrac{x-3}{x\left(2-x\right)}\)

\(\Leftrightarrow A=\left[\dfrac{4\left(x+2\right)}{\left(2-x\right)\left(2+x\right)}\right]:\dfrac{x-3}{x\left(2-x\right)}\)

\(\Leftrightarrow A=\dfrac{4}{2-x}.\dfrac{x\left(2-x\right)}{x-3}=\dfrac{4x}{x-3}\)

b) \(\left|x\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

- Với \(x=1:\)

\(A=\dfrac{4x}{x-3}=\dfrac{4.1}{1-3}=-2\)

- Với \(x=-1:\)

\(A=\dfrac{4x}{x-3}=\dfrac{4.\left(-1\right)}{-1-3}=1\)

Sự kiện kết thúc nhưng mà các bạn vẫn có thể góp ý cho sự kiện nhé!

Hay quá anh ơi

a) Ta có : 

\(BC^2=AB^2+AC^2\left(Pitago\right)\)

\(\Leftrightarrow AC^2=BC^2-AB^2=169-25=144\)

\(\Leftrightarrow AC=12\left(cm\right)\)

\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{AB^2.+AC^2}{AB^2.AC^2}\)

\(\Leftrightarrow\dfrac{1}{AH^2}=\dfrac{BC^2}{\left(AB.AC\right)^2}\)

\(\Leftrightarrow AH^2=\dfrac{\left(AB.AC\right)^2}{BC^2}=\dfrac{\left(5.12\right)^2}{13^2}\)

\(\Leftrightarrow AH=\dfrac{5.12}{13}=\dfrac{60}{13}\sim4,85\left(cm\right)\)

\(sin\widehat{B}=\dfrac{AC}{BC}=\dfrac{12}{13}\Rightarrow\widehat{B}\sim67^o\)

 a) ∆ABC vuông tại A (gt)

BC² = AB² + AC² (Pytago)A

⇒ AC² = BC² - AB²

= 13² - 5²

= 144

⇒ AC = 12 (cm)

Ta có:

AH.BC = AB.AC

⇒ AH = AB.AC : BC

= 5.12 : 13

= 60/13 (cm) ≈ 4,62 (cm)

sinB = AC/BC = 12/13

⇒ ∠B ≈ 67⁰

b) ∆AHB vuông tại H có HE là đường cao

⇒ HE² = AE . EB (1)

∆AHC vuông tại H có HF là đường cao

⇒ HF² = AF . FC (2)

Tứ giác AEHF có:

∠AEH = ∠EAF = ∠AFH = 90⁰

⇒ AEHF là hình chữ nhật

⇒ AH = EF

⇒ ∠EHF = 90⁰

∆EHF vuông tại H

⇒ EF² = HE² + HF²

⇒ AH² = HE² + HF²

Từ (1) và (2)

⇒ AE.EB + AF.FC = HE² + HF² = AH²

∆ABC vuông tại A vó AH là đường cao

⇒ AH² = HB.HC

⇒ AE.EB + AF.FC = HB.HC

⇒ AE.EB + AF.FC - HB.HC = 0

c) AH = EF đã chứng minh ở câu b

16 = 9 + 7 = 9 + √49

9 + 4√5 = 9 + √80

Do 49 < 80 nên √49 < √80

⇒ 9 + √49 < 9 + √80

Vậy 16 < 9 + 4√5

Ta có:

\(16=9+5=9+\sqrt{25}\)

\(9+4\sqrt{5}=9+\sqrt{4^2\cdot5}=9+\sqrt{20}\)

Mà: \(20< 25\)

\(\Rightarrow\sqrt{20}< \sqrt{25}\)

\(\Rightarrow9+\sqrt{20}< 9+\sqrt{25}\)

\(\Rightarrow9+4\sqrt{5}< 16\)

\(\sqrt[]{6+4\sqrt[]{2}}+\sqrt[]{11-6\sqrt[]{2}}\)

\(=\sqrt[]{4+2.2\sqrt[]{2}+2}+\sqrt[]{9-2.3\sqrt[]{2}+2}\)

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

\(=\left|2+\sqrt[]{2}\right|+\left|3-\sqrt[]{2}\right|\)

\(=2+\sqrt[]{2}+3-\sqrt[]{2}\left(3< \sqrt[]{2}\right)\)

\(=5\)