ta có :

\(B=\left(\frac{1}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right):\frac{x-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

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

ta có 

\(S\ge\frac{1}{\frac{1+1998}{2}}+\frac{1}{\frac{2+1997}{2}}+..+\frac{1}{\frac{k+1998-k+1}{2}}+..+\frac{1}{\frac{1999}{2}}\)

hay \(S\ge\frac{2}{1999}+\frac{2}{1999}+..+\frac{2}{1999}=2.\frac{1998}{1999}\)

do dấu = không xảy ra nên \(S>2.\frac{1998}{1999}\)

\(a,\sqrt{\left(3-\sqrt{5}\right)^2}+2\sqrt{5}=3-\sqrt{5}+2\sqrt{5}\)

\(=3+\sqrt{5}\)

\(b,\sqrt{9a^2}-\sqrt{\left(a-3\right)^2}\)

\(\sqrt{\left(3a\right)^2}-\left|a-3\right|\)

\(\left|3a\right|-a+3\)

\(2a+3\)

bài 5

\(a,\sqrt{\left(5-x\right)^2}=3\)

\(\left|5-x\right|=3\)

\(\orbr{\begin{cases}5-x=3\\5-x=-3\end{cases}\orbr{\begin{cases}x=2\left(TM\right)\\x=8\left(TM\right)\end{cases}}}\)

\(b,\sqrt{3x}-\sqrt{48x}+\sqrt{75x}=8\)

\(\sqrt{3x}\left(1-4+5\right)=8\)

\(\sqrt{3x}.0=8\left(KTM\right)\)

vậy pt vô nghiệm

\(\sqrt{x+1}+\sqrt{y+1}=4\Leftrightarrow x+y+2+2\sqrt{xy+x+y+1}=16\)

mà \(x+y=3+\sqrt{xy}\)nên ta có 

\(5+\sqrt{xy}+2\sqrt{xy+\sqrt{xy}+4}=16\Leftrightarrow2\sqrt{xy+\sqrt{xy}+4}=11-\sqrt{xy}\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{xy}\le11\\4\left(xy+\sqrt{xy}+4\right)=121-22\sqrt{xy}+xy\end{cases}\Leftrightarrow\hept{\begin{cases}\sqrt{xy}\le11\\3xy+26\sqrt{xy}-105=0\end{cases}}}\)

\(\Leftrightarrow\sqrt{xy}=3\Leftrightarrow\hept{\begin{cases}x+y=6\\xy=9\end{cases}}\Leftrightarrow x=y=3\)

a) Xét ΔABC và ΔHBA có

ˆBAC=ˆBHA(=900)BAC^=BHA^(=900)

ˆABHABH^ chung

Do đó: ΔABC∼ΔHBA(g-g)

b) Áp dụng định lí pytago vào ΔABC vuông tại A, ta được:

BC2=AB2+AC2BC2=AB2+AC2

⇔BC2=202+152=625⇔BC2=202+152=625

hay BC=√625=25cmBC=625=25cm

Ta có: ΔABC∼ΔHBA(cmt)

⇒ACHA=BCBAACHA=BCBA

hay 15AH=252015AH=2520

⇔AH=15⋅2025=30025=12cm⇔AH=15⋅2025=30025=12cm

Vậy: BC=25cm; AH=12cm

d) Ta có: ˆCAH+ˆBAH=ˆBACCAH^+BAH^=BAC^(tia AH nằm giữa hai tia AB,AC)

⇔ˆCAD=900−ˆBAHCAD^=900−BAH^(1)

Ta có: ΔAHB vuông tại H(AH⊥BC)

nên ˆABH+ˆBAH=900ABH^+BAH^=900(hai góc nhọn phụ nhau)

hay ˆABC=900−ˆBAHABC^=900−BAH^(2)

Từ (1) và (2) suy ra ˆCAD=ˆABCCAD^=ABC^

Ta có: CD//AB(gt)

AB⊥AC(ΔABC vuông tại A)

Do đó: CD⊥AC(định lí 2 từ vuông góc tới song song)

Xét ΔBAC và ΔACD có

ˆABC=ˆCADABC^=CAD^(cmt)

ˆBAC=ˆACD(=900)BAC^=ACD^(=900)

Do đó: ΔBAC∼ΔACD(g-g)

⇒ABAC=ACCDABAC=ACCD

hay AC2=AB⋅DCAC2=AB⋅DC(đpcm)

địt ko em

3.84 nha bạn đổi nãy giờ

\(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}=\sqrt[3]{2\sqrt{2}+6+3\sqrt{2}+1}-\sqrt[3]{2\sqrt{2}-6+3\sqrt{2}-1}\)

\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}-\sqrt[3]{\left(\sqrt{2}-1\right)^3}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)

suy ra \(a+b+c=2\)

Ta có: \(2^2=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)

\(\Leftrightarrow ab+bc+ca=\frac{4-\left(a^2+b^2+c^2\right)}{2}=\frac{4-1}{2}=\frac{3}{2}\)

\(\frac{3}{x-5}.\frac{\sqrt{\left(5-x\right)^3\left(x-1\right)}}{\sqrt{\left(x-1\right)^3}}-\frac{1}{x+1}=1\)

\(\frac{3}{x-5}.\frac{\left(x-5\right).-\sqrt{5-x}\sqrt{x-1}}{\left(x-1\right)\sqrt{x-1}}-\frac{1}{x+1}=1\)

\(\frac{-3\sqrt{5-x}}{\left(x-1\right)}-\frac{1}{x+1}=1\)

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

\(\frac{-3x\sqrt{5-x}-3\sqrt{5-x}+1}{x^2-1}=1\)

\(-3x\sqrt{5-x}-3\sqrt{5-x}+1=x^2-1\)

\(\left(-3x\sqrt{5-x}-3\sqrt{5-x}\right)^2=\left(x^2-2\right)^2\)

\(9x^2\left(5-x\right)-9\left(5-x\right)=x^4-4x^2+4\)

\(45x^2-9x^3-45+9x=x^4-4x^2+4\)

\(x^4+9x^3-49x^2-9x+49=0\)

\(x^4+9x^3=49x^2+9x-49\)

\(\left(x^2\right)^2+x^2.\frac{9}{2}x+\frac{81}{4}x^2=49x^2+9x-49+\frac{81}{4}x^2\)

\(\left(x^2+\frac{9x}{2}\right)^2=49x^2+9x-49+\frac{81}{4}x^2\)

\(\left(x^2+\frac{9x}{2}\right)^2+\left(x^2+\frac{9x}{2}\right)y+\frac{y^2}{4}=49x^2+9x-49+\frac{81}{4}x^2+\left(x^2+\frac{9}{2}\right)y+\frac{y^2}{4}\)

\(\left(x^2+\frac{9}{2}+\frac{y}{2}\right)^2=\frac{277}{4}x^2+9x-49+\left(x^2+\frac{9}{2}x\right)y+\frac{y^2}{4}\)

\(\left(x^2+\frac{9x}{2}+\frac{y}{2}\right)^2=x^2\left(\frac{277}{4}+y\right)+x\left(9+\frac{9}{2}y\right)+\frac{y^2}{4}-49\)

đặt pt vế phải là f(x) tính delta

\(\Delta=\left(9+\frac{9}{2}y\right)^2-4\left(\frac{277}{4}+y\right)\left(\frac{y^2}{4}-49\right)\)

\(\Delta=81+81y+\frac{81}{4}y^2-\left(277+4y\right)\left(\frac{y^2}{4}-49\right)\)

\(\Delta=81+81y+\frac{81}{4}y^2-\frac{277}{4}y^2-y^3+13573+196y\)

\(\Delta=13654+277y-49y^2-y^3\)

bạn giải pt bậc ba ra xong thế vào pt trên thì ra đc 4 nghiệm của pt

bạn tự làm nốt nhé