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giải giúp mình hệ phương trình 3 ẩn này với mình cảm ơn nhiều
3=(a+b)(a+c)
4=(a+b)(b+c)
5=(a+c)(b+c)
Đáp án: C
A ∩ B = {b; d}; A ∩ C = {a; b}; B ∩ C = {b; e}
A \ B = {a; c}; A \ C = {c; d}; B \ C = {d}
A ∪ B = {a; b; c; d; e}; A ∪ C = {a; b; c; d; e}
A ∩ (B \ C) = {d}. (A ∩ B) \ (A ∩ C) = {d}.
A \ (B ∩ C) = {a; c; d}. (A \ B) ∪ (A \ C) = {a; c; d}.
(A \ B) ∩ (A \ C) = {c}.
a. A ∩ (B \ C) = (A ∩ B) \ (A ∩ C) ={d} ⇒ a đúng.
b. A \ (B ∩ C)= {a; c; d} (A \ B) ∩ (A \ C)={c} ⇒ b sai.
c. A ∩ (B \ C) ={d} (A \ B) ∩ (A \ C)={c} ⇒ c sai
d. A \ (B ∩C) = (A \ B) ∪ (A \ C)= {a; c; d} ⇒ d đúng.
\(S=\dfrac{b+c}{a}+\dfrac{c+a}{b}+\dfrac{a+b}{c}\)
\(S=\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c}=a\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{a}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge a.\dfrac{4}{b+c}+b.\dfrac{4}{a+c}+c.\dfrac{4}{a+b}=4\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)\)
Dạ vâng ạ, em chiều nay cũng vừa nghĩ ra được cách này.
Em cám ơn nhiều lắm ạ!
mình nghĩ đề nó như thế này
\(\sqrt{a^2+b^2}-\sqrt{c^2+d^2}\ge\sqrt{\left(a+c\right)^2-\left(b+d^{ }\right)^2}\)
hai zế BĐT ko âm nên bình phương 2 zế ta có
\(a^2+b^2+c^2+d^2+2\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge a^2+2ac+c^2+b^2+2bd+d^2\)
\(\Leftrightarrow\sqrt{\left(a^2+b^2\right)\left(c^2+d^2\right)}\ge ac+bd\left(1\right)\)
Nếu \(ac+bd< 0\)thì BĐT đc c/m
Nêu \(ac+bd\ge0\left(1\right)\Leftrightarrow\left(a^2+b^2\right)\left(c^2+d^2\right)\ge a^2c^2+b^2d^2+2acbd\)
\(\Leftrightarrow a^2c^2+a^2d^2+b^2c^2+b^2d^2\ge a^2c^2+b^2d^2+2acbd\)
\(\Leftrightarrow a^2d^2+b^2c^2-2acbd\ge0\Leftrightarrow\left(ad-bc\right)^2\ge0\)( luôn đúng )
dấu = xảy ra khi \(ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\)
Bài 1:
\(\left(2x+3\right)^2-\left(2x+3\right)\left(4x-6\right)+\left(2x-3\right)^2+xy\)
\(=\left(2x+3\right)^2-2\cdot\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2+xy\)
\(=\left(2x+3-2x+3\right)^2+xy\)
\(=xy+36=2\cdot\left(-1\right)+36=36-2=34\)
Bài 2:
a: \(a^2+b^2+c^2\ge ab+bc+ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(a^2-2ac+c^2\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng)
b: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left(a^2+2ab+b^2+2ac+c^2+2bc+a^2+ab+ac+a^2-b^2+bc-c^2\right)\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
a)\(-\left(a-c\right)-\left(a-b+c\right)\)
\(=-a+c-a+b-c\)
\(=\left(-a-a\right)+\left(c-c\right)+b\)
\(=-2a+b=0\)
\(-2a=b\)
b)\(-\left(a-b+c\right)-\left(a+b+c\right)\)
\(=-a+b-c-a-b-c\)
\(=\left(-a-a\right)+\left(b-b\right)+\left(-c-c\right)\)
\(=-2a+0=-2c\)
\(=-2a+-2c\)
c)\(\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)\)
\(=a+b-a+b+a-c-a-c\)
\(=\left(a-a+a-a\right)+\left(b+b\right)+\left(-c-c\right)\)
\(=0+2b+\left(-2c\right)\)
\(=2b+\left(-2c\right)\)
d)\(\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(=a+b-c+a-b+c-b-c+a-a+b+c\)
\(=\left(a+a+a-a\right)+\left(b-b-b+a\right)+\left(-c+c-c+c\right)\)
\(=2a+0+0\)
\(=2a\)
a)-(a-c)-(a-b+c)
=-a+c-a+b-c
=-2a+b
b)-(a-b+c)-(a+b+c)
=-a+b-c-a-b-c
=-2a-2c
=-2(a+c)
c)(a+b)-(a-b)+(a-c)-(a+c)
=a+b-a+b+a-c-a-c
=2b-2c
=2(b-c)
d)(a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)
=a+b-c+a-b+c-b-c+a-a+b+c
=2a