Phương An

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Tam giác HAB có HD là đường cao

\(\Rightarrow AH^2=AD\times AB\left(htl\right)\left(1\right)\)

Tam giác HAC có HE là đường cao

\(\Rightarrow AH^2=AE\times AC\left(htl\right)\left(2\right)\)

(1) và (2) => đpcm

~ ~ ~ ~ ~

HDA = DAE = AEH = 90⁰

=> ADHE là hcn

=> EDH = AHD và HED = EHA

- - -

Tam giác DBH vuông tại D có DM là trung tuyến (M là trung điểm của BH)

=> DM = MH

=> Tam giác MDH cân tại M

=> MDH = MHD

Ta có: MDE = MDH + HDE = MHD + DHA = AHB = 90⁰

=> MD _I_ DE

=> DE là tiếp tuyến của đường tròn (M ; MD) (3)

- - -

Tam giác ECH vuông tại E có EN là trung tuyến (N là trung điểm của CH)

=> EN = NH

=> Tam giác NEH cân tại N

=> NEH = NHE

Ta có: NED = NEH + HED = NHE + EHA = AHC = 90⁰

=> NE _I_ DE

=> DE là tiếp tuyến của đường tròn (N ; NE) (4)

(3) và (4) => đpcm

~ ~ ~ ~ ~

Tam giác ABC vuông tại A có AH là đường cao:

(+) BC² = AB² + AC² (ptg)

=> BC = 10 (cm)

(+) AB² = BH . BC (htl)

=> BH = 3,6 (cm)

(+) AC² = HC . BC (htl)

=> HC = 6,4 (cm)

\(DM=\dfrac{BH}{2}=1,8\left(cm\right)\)

\(EN=\dfrac{HC}{2}=3,2\left(cm\right)\)

MD _I_ DE và NE _I_ ED

=> MD // NE

=> MDEN là hình thang

Q là trung điểm của DE (ADHE là hcn)

P là trung điểm của MN (gt)

=> PQ là đtb của hình thang MDEN

\(\Rightarrow PQ=\dfrac{\left(DM+EN\right)}{2}=2,5\left(cm\right)\)

~ ~ ~ ~ ~

Áp dụng BĐT AM - GM, ta có:

\(\dfrac{x^4+5}{\sqrt{x^4+4}}=\dfrac{x^4+4+1}{\sqrt{x^4+4}}=\sqrt{x^4+4}+\dfrac{1}{\sqrt{x^4+4}}\ge2\)

Dấu "=" xảy ra khi \(x^4=-3\) (vô lý)

Vậy \(\dfrac{x^4+5}{\sqrt{x^4+4}}>2\forall x\)

\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{8-6\sqrt{x-1}+x}=1\)

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

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

Ta có:

\(VT=\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)

mà VP = 1

Vậy pt vô nghiệm.

\(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}}=1\)

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

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

Ta có:

\(VT=\left|\sqrt{x-2}-2\right|+\left|3-\sqrt{x-2}\right|\ge\left|\sqrt{x-2}-2+3-\sqrt{x-2}\right|=1\)

Dấu "=" xảy ra khi \(\left(\sqrt{x-2}-2\right)\left(3-\sqrt{x-2}\right)\ge0\)

Bảng xét dấu:

Vậy \(6\le x\le11\)

\(x^2-4x-6=\sqrt{2x^2-8x+12}\)

\(\Leftrightarrow\left(x^2+2x\right)-\left(6x+6+\sqrt{2x^2-8x+12}\right)=0\)

\(\Leftrightarrow x\left(x+2\right)-\dfrac{36x^2+72x+36-\left(2x^2-8x+12\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)

\(\Leftrightarrow x\left(x+2\right)-\dfrac{2\left(17x+6\right)\left(x+2\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}=0\)

\(\Leftrightarrow\left(x+2\right)\left[x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\right]=0\)

Pt \(x-\dfrac{2\left(17x+6\right)}{\left(6x+6\right)-\sqrt{2x^2-8x+12}}\) vô nghiệm

=> x + 2 = 0

<=> x = - 2 (nhận)

\(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{25}\)

\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+20\)

\(=20-11\sqrt{5}\)

~ ~ ~ ~ ~

\(\sqrt{8x}-\sqrt{18x}+2\sqrt{32x}=14\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{2}-3\sqrt{2}+8\sqrt{2}\right)=14\)

\(\Leftrightarrow\sqrt{x}\times7\sqrt{2}=14\)

\(\Leftrightarrow98x=196\)

\(\Leftrightarrow x=2\) (nhận)

\(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\sqrt{x}+1}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{7-6\sqrt{x}}{\sqrt{x}+1}\)

\(=\dfrac{\left(x+\sqrt{x}\right)+\left(7\sqrt{x}-7-6x+6\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{-5x+14\sqrt{x}-7}{x-1}\)

\(\dfrac{-5x+14\sqrt{x}-7}{x-1}< \dfrac{1}{2}\)

\(\Leftrightarrow2\left(-5x+14\sqrt{x}-7\right)< x-1\)

\(\Leftrightarrow11x-28\sqrt{x}+13>0\)

x² + 14x + y² - 12y + 47

= x² + 14x + 49 + y² - 12y + 36 - 38

= (x + 7)² + (y - 6)² - 38 \(\ge\) - 38

Dấu "=" xảy ra khi x = - 7 và y = 6

x² + 2x - 8y² = 41

<=> (x + 1)² - 8y² = 42 (1)

Ta có:

\(-8y^2⋮2\) và \(42⋮2\) \(\Rightarrow\left(x+1\right)^2⋮2\Rightarrow x⋮̸2\)

Đặt x = 2k + 1 \(\left(k\in Z\right)\), ta có:

\(\left(1\right)\Leftrightarrow\left(2k+1+1\right)^2-8y^2=42\)

\(\Leftrightarrow4\left(k+1\right)^2-8y^2=42\)

\(\Leftrightarrow2\left(k+1\right)^2-4y^2=21\)

Ta có:

\(VT⋮2\) mà \(VP⋮̸2\)

Vậy pt không có nghiệm nguyên.

\(\dfrac{HB}{HC}=\dfrac{1}{4}\left(gt\right)\Rightarrow HC=4HB\)

Tam giác ABC vuông tại A có AH là đường cao:

\(AH^2=HB\times HC=HB\times4HB=4HB^2\left(htl\right)\)

\(\Rightarrow HB=7\left(cm\right)\)

\(\Rightarrow HC=4\times7=28\left(cm\right)\)

\(S_{ABC}=S_{HAC}+S_{HAB}=\dfrac{1}{2}\times AH\times\left(HB+HC\right)=1372\left(cm^2\right)\)

Tam giác HAC vuông tại H có:

\(AC^2=AH^2+HC^2\left(ptg\right)\)

\(\Rightarrow AC=14\sqrt{5}\left(cm\right)\)

Tam giác HAB vuông tại H có:

\(AB^2=AH^2+BH^2\left(ptg\right)\)

\(\Rightarrow AB=7\sqrt{5}\left(cm\right)\)

\(P_{ABC}=AB+AC+BC=14\sqrt{5}+7\sqrt{5}+\left(HB+HC\right)=42+21\sqrt{5}\left(cm\right)\)

Tam giác ABD vuông tại A có AO là đường cao:

\(\dfrac{1}{AO^2}=\dfrac{1}{AB^2}+\dfrac{1}{AD^2}\left(hlt\right)\)

\(\Rightarrow AO=12\left(cm\right)\)

Tam giác OAB vuông tại O:

\(AB^2=OA^2+OB^2\left(ptg\right)\)

\(\Rightarrow OB=9\left(cm\right)\)

Tam giác OAD vuông tại O:

\(AD^2=OA^2+OD^2\left(ptg\right)\)

\(\Rightarrow OD=16\left(cm\right)\)

Tam giác DAC vuông tại D có DO là đường cao:

\(AD^2=OA\times AC\left(htl\right)\)

\(\Rightarrow AC=\dfrac{100}{3}\left(cm\right)\)

Tam giác DAC vuông tại D có:

\(AC^2=AD^2+DC^2\left(ptg\right)\)

\(\Rightarrow DC=\dfrac{80}{3}\left(cm\right)\)

\(S_{ABCD}=\dfrac{1}{2}\times\left(AB+CD\right)\times AD=\dfrac{1250}{3}\left(cm^2\right)\)

\(P=\dfrac{\sqrt{x}+1}{x-1}-\dfrac{x-1}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)

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

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

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

\(=\dfrac{1}{\sqrt{x}-1}\)
~ ~ ~

\(2\div\dfrac{1}{\sqrt{x}-1}+\sqrt{x}\)

\(=2\sqrt{x}+\sqrt{x}-2\)

\(=3\sqrt{x}-2\ge-2\)

Dấu "=" xảy ra khi x = 0

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

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

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

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

Dấu "=" xảy ra khi x = 0,25

~ ~ ~

\(Q_1=\dfrac{3\left(x-\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{3}{\sqrt{x}}=3\sqrt{x}-3+\dfrac{3}{\sqrt{x}}\)

Q₁ đạt giá trị nguyên

<=> \(\sqrt{x}\in\text{Ư}\left(3\right)=\left\{1;3\right\}\Leftrightarrow x\in\left\{1;9\right\}\)

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

\(=\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)\(\div\left[\dfrac{\left(\sqrt{x}-1\right)-\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\)

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

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

~ ~ ~

\(-\dfrac{2}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{8}\ge1\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{8}\le-1\)

\(\Leftrightarrow16+x+2\sqrt{x}+1\le-8\sqrt{x}-8\)

\(\Leftrightarrow x+10\sqrt{x}+25\le0\)

\(\Leftrightarrow\left(\sqrt{x}+5\right)^2\le0\)

\(\Leftrightarrow\sqrt{x}\le-5\) (vô lý)

Vậy không có giá trị nào của x thoả mãn yêu cầu.

Sửa đề: Chứng minh \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)

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Áp dụng BĐT Cauchy Shwarz, ta có:

\(\left(1+1+1\right)\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2\ge\dfrac{\left(a+b+c\right)^4}{9}\)

Áp dụng BĐT Cauchy Shwarz và BĐT AM - GM, ta có:

(+) \(\left(1+1+1\right)\left(a^4+b^4+c^4\right)\ge\left(a^2+b^2+c^2\right)^2\)

\(\Leftrightarrow a^4+b^4+c^4\ge\dfrac{\left(a^2+b^2+c^2\right)^2}{3}\)

\(\ge\dfrac{\dfrac{\left(a+b+c\right)^4}{9}}{3}=\dfrac{\left(a+b+c\right)^3}{27}\times\left(a+b+c\right)\)

\(\ge\dfrac{\left(3\sqrt[3]{abc}\right)^3}{27}\times\left(a+b+c\right)=abc\left(a+b+c\right)\)

Dấu "=" xảy ra khi a = b = c

b)

M = - x² - x - y² - 3y + 13

4M = - 4x² - 4x - 4y² - 12y + 52

= - (2x + 1)² - (2y + 3)² + 42 \(\le\) 42

\(M\le\dfrac{21}{2}\)

Dấu "=" xảy ra khi \(x=-\dfrac{1}{2}\) và \(y=-\dfrac{3}{2}\)

a) Biến đổi tương đương:

\(a^2+b^2+1\ge ab+a+b\) (1)

\(\Leftrightarrow2a^2+2b^2+2\ge2ab+2a+2b\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\) luôn đúng

=> (1) đúng

Dấu "=" xảy ra khi a = b = 1

b) Áp dụng BĐT Cauchy Shwarz dạng Engel, ta có:\(\dfrac{a^2+b^2+c^2}{3}=\dfrac{a^2}{3}+\dfrac{b^2}{3}+\dfrac{c^2}{3}\ge\dfrac{\left(a+b+c\right)^2}{3+3+3}=\left(\dfrac{a+b+c}{3}\right)^2\)

Dấu "=" xảy ra khi a = b = c.

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

\(A\in Z\Leftrightarrow\dfrac{2}{x+1}\in Z\Leftrightarrow x+1\in\text{Ư}\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(\Leftrightarrow x\in\left\{-3;-2;0;1\right\}\)

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

\(B\in Z\Leftrightarrow\dfrac{3}{\sqrt{x}-3}\in Z\Leftrightarrow\sqrt{x}-3\in\text{Ư}\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(\Leftrightarrow x\in\left\{0;3\right\}\)

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

\(\Leftrightarrow\sqrt{2x-3}-\sqrt{x-1}-\sqrt{x-2}=0\)

\(\Leftrightarrow\dfrac{2x-3-x+1}{\sqrt{2x-3}+\sqrt{x-1}}-\sqrt{x-2}=0\)

\(\Leftrightarrow\dfrac{x-2}{\sqrt{2x-3}+\sqrt{x-1}}-\dfrac{x-2}{\sqrt{x-2}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{\sqrt{2x-3}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x-2}}\right)=0\)

Pt \(\dfrac{1}{\sqrt{2x-3}+\sqrt{x-1}}-\dfrac{1}{\sqrt{x-2}}=0\) vô nghiệm

=> x - 2 = 0

<=> x = 2 (nhận)

\(\sqrt{3x+7}-\sqrt{x+1}=2\)

\(\Leftrightarrow\sqrt{3x+7}-2-\sqrt{x+1}=0\)

\(\Leftrightarrow\dfrac{3x+7-4}{\sqrt{3x+7}+2}-\dfrac{x+1}{\sqrt{x+1}}=0\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)}{\sqrt{3x+7}+2}-\dfrac{x+1}{\sqrt{x+1}}=0\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{3}{\sqrt{3x+7}+2}-\dfrac{1}{\sqrt{x+1}}\right)=0\)

Pt \(\dfrac{3}{\sqrt{3x+7}+2}-\dfrac{1}{\sqrt{x+1}}=0\) vô nghiệm

=> x + 1 = 0

<=> x = - 1 (nhận)

\(M=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\)

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

\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)

\(=x-2\sqrt{x}+1\)

\(\sqrt{2x+3}+\sqrt{2x+2}=1\)

\(\Leftrightarrow\sqrt{2x+3}-1+\sqrt{2x+2}=0\)

\(\Leftrightarrow\dfrac{2x+3-1}{\sqrt{2x+3}-1}+\dfrac{2x+2}{\sqrt{2x+2}}=0\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)}{\sqrt{2x+3}-1}+\dfrac{2\left(x+1\right)}{\sqrt{2x+2}}=0\)

\(\Leftrightarrow2\left(x+1\right)\left(\dfrac{1}{\sqrt{2x+3}-1}+\dfrac{1}{\sqrt{2x+2}}\right)=0\)

Pt \(\dfrac{1}{\sqrt{2x+3}-1}+\dfrac{1}{\sqrt{2x+2}}=0\) vô nghiệm

=> x + 1 = 0

<=> x = -1

\(\sqrt{x+4}-\sqrt{2x-6}=1\)

\(\Leftrightarrow\sqrt{x+4}-3-\sqrt{2x-6}+2=0\)

\(\Leftrightarrow\dfrac{x+4-9}{\sqrt{x+4}+3}-\dfrac{2x-6-4}{\sqrt{2x-6}+2}=0\)

\(\Leftrightarrow\dfrac{x-5}{\sqrt{x+4}+3}-\dfrac{2\left(x-5\right)}{\sqrt{2x-6}+2}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{1}{\sqrt{x+4}+3}-\dfrac{2}{\sqrt{2x-6}+2}\right)=0\)

Pt \(\dfrac{1}{\sqrt{x+4}+3}-\dfrac{2}{\sqrt{2x-6}+2}=0\) vô nghiệm

=> x - 5 = 0

<=> x = 5 (nhận)

\(\sqrt{1-x}+\sqrt{4+x}=3\)

\(\Leftrightarrow\sqrt{1-x}-1+\sqrt{4+x}-2=0\)

\(\Leftrightarrow\dfrac{1-x-1}{\sqrt{1-x}+1}+\dfrac{4+x-4}{\sqrt{4+x}+\sqrt{2}}=0\)

\(\Leftrightarrow\dfrac{x}{\sqrt{4+x}+\sqrt{2}}-\dfrac{x}{\sqrt{1-x}+1}=0\)

\(\Leftrightarrow x\left(\dfrac{1}{\sqrt{4+x}+\sqrt{2}}-\dfrac{1}{\sqrt{1-x}+1}\right)=0\)

Pt \(\dfrac{1}{\sqrt{4+x}+\sqrt{2}}-\dfrac{1}{\sqrt{1-x}+1}=0\) vô nghiệm

=> x = 0