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4) a) Giả sử \(a\ge b\ge c\). Ta có
\(a< b+c\Leftrightarrow2a< a+b+c=2\Rightarrow a< 1\Rightarrow b< 1;c< 1\)
b) Từ câu a), ta suy ra: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)
\(\Leftrightarrow ab+bc+bc>1+abc\)
Mặt khác, ta cũng có: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)\)
\(\Rightarrow4>a^2+b^2+c^2+2\left(1+abc\right)\)
\(\Rightarrow4>a^2+b^2+c^2+2+2abc\)
\(\Rightarrow2-2abc>a^2+b^2+c^2\Rightarrow a^2+b^2+c^2< 2\left(1-abc\right)\)
MK hứng bài nào thì lm bài đấy nhé!
Bài 21:
Ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
<=> \(\dfrac{ab+bc+ca}{abc}=0\)
<=> \(ab+bc+ac=0\)
<=> \(ab+bc+ac+c^2=c^2\)
<=> \(\sqrt{ab+bc+ac+c^2}=\sqrt{c^2}\)
<=> \(\sqrt{\left(a+c\right)\left(b+c\right)}=\left|c\right|\) (1)
Mặt khác: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) ; \(a,b>0;c\ne0\) => \(c< 0\) (2)
Từ (1); (2) => \(\sqrt{\left(a+c\right)\left(b+c\right)}=-c\)
<=> \(2\sqrt{\left(a+c\right)\left(b+c\right)}+2c=0\)
<=> \(\left(a+c\right)+2\sqrt{\left(a+c\right)\left(b+c\right)}+\left(b+c\right)=a+b\)
<=> \(\left(\sqrt{a+c}+\sqrt{b+c}\right)^2=\left(\sqrt{a+b}\right)^2\)
<=> \(\sqrt{a+c}+\sqrt{b+c}=\sqrt{a+b}\) => Đpcm
Bài 2 :
a ) \(\sqrt{4x-8}+\sqrt{x-2}=4+\dfrac{1}{3}\sqrt{9x-18}\) ( ĐKXĐ : \(x\ge2\) )
\(\Leftrightarrow2\sqrt{x-2}+\sqrt{x-2}=4+\dfrac{1}{3}.3\sqrt{x-2}\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=4\)
\(\Leftrightarrow2\sqrt{x-2}=4\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\)
\(\Leftrightarrow x=2\) ( thỏa mãn ĐKXĐ )
Vậy phương trình có nghiệm x = 2 .
Bài 2 :
b ) \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}-\dfrac{\sqrt{3}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=0\)
\(\Leftrightarrow|x-3|-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{3}=0\left(x\ge3\right)\\3-x-\sqrt{3}=0\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{3}\\x=3-\sqrt{3}\end{matrix}\right.\)
Vậy phương trình cón nghiệm \(x=3+\sqrt{3}\) hoặc \(x=3-\sqrt{3}\) .
Bạn đúng là 1 người tốt bụng , quan tâm tới bạn bè , chắc chắn mọi điều tốt sẽ đến vs bạn
Mặc dù mk ko bt bạn Hạ Thì là aiNNhưng mk chúc mừng sinh nhật bạn ấy







please help me
vs

thì g


giúp vs ạ




thì e chúc sau



















1 điểm bằng mấy k bạn ơi
dương lý khánh hạ là sao bạn ???
dfghjkl
Câu 5:
Hình đa giác TenDaGiac1: DaGiac(B, C, 4) Góc α: Góc giữa G, O, G' Góc α: Góc giữa G, O, G' Đoạn thẳng f: Đoạn thẳng [B, C] của Hình đa giác TenDaGiac1 Đoạn thẳng g: Đoạn thẳng [C, D] của Hình đa giác TenDaGiac1 Đoạn thẳng h: Đoạn thẳng [D, A] của Hình đa giác TenDaGiac1 Đoạn thẳng i: Đoạn thẳng [A, B] của Hình đa giác TenDaGiac1 Đoạn thẳng j: Đoạn thẳng [A, C] Đoạn thẳng k: Đoạn thẳng [B, D] Đoạn thẳng m: Đoạn thẳng [G, O] Đoạn thẳng n: Đoạn thẳng [O, H] Đoạn thẳng p: Đoạn thẳng [M, G] Đoạn thẳng q: Đoạn thẳng [A, H] Đoạn thẳng r: Đoạn thẳng [M, O] B = (12.15, -29.36) B = (12.15, -29.36) B = (12.15, -29.36) C = (52.4, -29.36) C = (52.4, -29.36) C = (52.4, -29.36) Điểm D: DaGiac(B, C, 4) Điểm D: DaGiac(B, C, 4) Điểm D: DaGiac(B, C, 4) Điểm A: DaGiac(B, C, 4) Điểm A: DaGiac(B, C, 4) Điểm A: DaGiac(B, C, 4) Điểm O: Giao điểm đường của j, k Điểm O: Giao điểm đường của j, k Điểm O: Giao điểm đường của j, k Điểm M: Trung điểm của i Điểm M: Trung điểm của i Điểm M: Trung điểm của i Điểm G: Điểm trên f Điểm G: Điểm trên f Điểm G: Điểm trên f Điểm H: Giao điểm đường của l, g Điểm H: Giao điểm đường của l, g Điểm H: Giao điểm đường của l, g
a) Do \(\widehat{HOG}=45^o\Rightarrow\widehat{BOG}+\widehat{DOH}=180^o-45^o=135^o\)
Xét tam giác DOH có \(\widehat{ODH}=45^o\Rightarrow\widehat{DHO}+\widehat{DOH}=135^o\)
Vậy nên \(\widehat{DHO}=\widehat{BOG}\)
Xét tam giác HOD và OGB có :
\(\widehat{ODH}=\widehat{GBO}=45^o\)
\(\widehat{DHO}=\widehat{BOG}\left(cmt\right)\)
Vậy nên \(\Delta HOD\sim\Delta OGB\left(g-g\right)\)
b) Do \(\Delta HOD\sim\Delta OGB\Rightarrow\frac{HD}{OB}=\frac{OD}{GB}\)
Đặt MB = x thì AD = 2x. Vậy \(OB=OD=x\sqrt{2}\)
Vậy thì \(HD.GB=OB.OD=x\sqrt{2}.x\sqrt{2}=2x^2=AD.BM\)
Từ đó suy ra \(HD.GB=AD.BM\Rightarrow\frac{HD}{MB}=\frac{AD}{GB}\)
Vậy ta có ngay \(\Delta ADH\sim\Delta GBM\left(c-g-c\right)\Rightarrow\widehat{AHD}=\widehat{GMB}\Rightarrow\widehat{HAB}=\widehat{GMB}\)
Chúng lại ở vị trí đồng vị nên MG // AH.
Bài 6:
A B C D I
Trên tia đối của tia DA, lấy điểm I sao cho \(\widehat{DCI}=\widehat{BAD}=\widehat{DAC}\)
Vậy thì \(\Delta BAD\sim\Delta ICD\left(g-g\right)\Rightarrow\widehat{ABD}=\widehat{CID};\frac{BD}{ID}=\frac{AD}{CD}\Rightarrow BD.CD=AD.ID\)
Từ đó ta cũng có \(\Delta BAD\sim\Delta IAC\left(g-g\right)\Rightarrow\frac{BA}{IA}=\frac{AD}{AC}\Rightarrow AB.AC=AI.AD\)
Từ đây \(AB.AC-BD.BC=AD.AI-AD.ID=AD\left(AI-ID\right)=AD^2\)
Vậy \(AD=\sqrt{AB.AC-BD.DC}\left(đpcm\right)\)
Bài 4:
a) Ta có a + b + c = 2.
+ Nếu cả ba cạnh lớn hơn 1 thì chu vị lớn hơn 3 (Loại)
+ Nếu chỉ tồn tại hai cạnh lớn hơn 1 thì chu vi lớn hơn 2 (Loại)
+ Nếu chỉ tồn tại một cạnh lớn hơn 1 thì tổng hai cạnh còn lại sẽ phải nhỏ hơn 1. Điều này vô lí vì theo bất đẳng thức trong tam giác, độ dài một cạnh phải nhỏ hơn tổng độ dài hai cạnh còn lại.
Tóm lại cả a, b và c đều phải nhỏ hơn 1.
b) Đề bài tương đương: Cho \(0< a,b,c< 1\). Chứng minh rằng \(a^2+b^2+c^2< 2\left(1-abc\right)\)
Ta có \(S=\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)};p=2:2=1\)
\(\Rightarrow S^2=\left(1-a\right)\left(1-b\right)\left(1-c\right)=1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc\)
\(=-1+\left(ab+bc+ca\right)-abc>0\)
\(\Rightarrow abc+1< ab+bc+ca\)
Mà \(ab+bc+ca=\frac{\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)}{2}=\frac{4-\left(a^2+b^2+c^2\right)}{2}\)
\(\Rightarrow abc+1< \frac{4-\left(a^2+b^2+c^2\right)}{2}\Rightarrow a^2+b^2+c^2< 2-2abc=2\left(1-abc\right)\left(đpcm\right)\)
a/ Ta có:
\(a+a< a+b+c=2\)
\(\Leftrightarrow a< 1\)
Tương tự ta có \(\hept{\begin{cases}b< 1\\c< 1\end{cases}}\)
b/ Ta có: \(\left(1-a\right)\left(1-b\right)\left(1-c\right)>0\)
\(\Leftrightarrow1-\left(a+b+c\right)+\left(ab+bc+ca\right)-abc>0\)
\(\Leftrightarrow-1+\left(ab+bc+ca\right)-abc>0\)
\(\Leftrightarrow-2+2\left(ab+bc+ca\right)-2abc>0\)
\(\Leftrightarrow-2+a^2+b^2+c^2+2\left(ab+bc+ca\right)-2abc>a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2< 2-2abc=2\left(1-abc\right)\)
7/ Ta có:
Thay \(x=9\) vào ta được
\(f\left(\sqrt{9}-1\right)+f\left(2\right)=9^2\)
\(\Leftrightarrow f\left(2\right)+f\left(2\right)=81\)
\(\Leftrightarrow f\left(2\right)=\frac{81}{2}\)
Thay \(x=\left(\sqrt{2}+1\right)^2\) vào ta được
\(f\left(\sqrt{\left(\sqrt{2}+1\right)^2}-1\right)+f\left(2\right)=\left(\sqrt{2}+1\right)^4\)
\(\Leftrightarrow f\left(\sqrt{2}\right)+f\left(2\right)=\left(\sqrt{2}+1\right)^4\)
\(\Leftrightarrow f\left(\sqrt{2}\right)=\left(\sqrt{2}+1\right)^4-f\left(2\right)=\left(\sqrt{2}+1\right)^4-\frac{81}{2}\)
cô nhầm hay sao ạ
BĐT tam giác là tổng 2 cạnh ;ớn hơn hoặc bằng cạnh còn lại chứ ạ
a+b\(\ge c\)suy ra c\(\le1\)
a+c\(\ge\)b suy ra b\(\le1\)
b+c\(\ge a\)suy ra a \(\le1\)
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