Cho a-b-c=2
Tính M=\(\frac{a^3-b^3-c^3-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
Thanks các bạn!
Cho a-b-c=2
Tính M=\(\frac{a^3-b^3-c^3-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
Giúp mình nhé mình đang cần gấp. Thanks các bạn
\(=\frac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{a^2+2ab+b^2+b^2-2bc+c^2+c^2+2ca+a^2}\)
\(=\frac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2\right)+3ab\left(a-b-c\right)}{\left(a-b-c\right)^2+a^2+b^2+c^2}\)
\(=\frac{\left(\cdot a-b-c\right)\left(a^2+b^2+c^2+ac+ab-bc\right)}{4+a^2+b^2+c^2}\)
\(=\frac{2a^2+2b^2+2c^2+2ab-2bc+2ca}{4+a^2+b^2+c^2}\)
\(=\frac{\left(a-b-c\right)^2+a^2+b^2+c^2}{4+a^2+b^2+c^2}=1\)
k mk nha
Cho a-b-c=2
Tính M=\(\frac{a^3-b^3-c^3-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
Giúp mình nhé mình đang cần gấp. Thanks các bạn
\(M=\dfrac{\left(a-b\right)^3-c^3+3ab\left(a-b\right)-3abc}{\left(a+b\right)^2+\left(b-c\right)^2+\left(c+a\right)^2}\)
\(=\dfrac{\left(a-b-c\right)\left(a^2-2ab+b^2+ac-bc+c^2+3ab\right)}{2a^2+2b^2+2c^2+2ab-2bc+2ac}\)
\(=\dfrac{\left(a-b-c\right)\cdot\left(a^2+b^2+c^2-ab-bc+ac\right)}{2\cdot\left(a^2+b^2+c^2+ab-bc+ac\right)}=\dfrac{2}{2}=1\)
Rút gọn biểu thức
a.\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
b.\(\frac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
c.\(\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
d.\(\left(x^2-x+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right)\left(x^{16}-x^8+1\right)\)
MONG CÁC BẠN CÓ THỂ BỎ RA VÀI PHÚT ĐỂ GIÚP MÌNH=))NÓ CŨNG GIÚP BẠN ÔN TẬP ĐƯỢC CÁC BÀI CHUẨN BỊ CHO KÌ THÌ MÀ=))MÌNH XIN CẢM ƠN RẤT RẤT NHIỀU
a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)
\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)
\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)
\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)
\(=\frac{a-b}{b+c}\)
cho a a+ b + c = 5
rút gọn
P = \(\frac{a^3+b^3+c^3-3abc}{\left(a-b^2\right)+\left(b-c^2\right)+\left(c-a^2\right)}\)
ai làm được tick cho
thanks
Sửa đề: \(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(P=\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
\(P=\frac{\left(a+b\right)^3+c^3-3abc-3a^2b-3ab^2}{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2}\)
\(P=\frac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right).c+c^2\right]-3ab\left(a+b+c\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(P=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2+2ab-ac-bc+3ab\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)
\(P=\frac{5\left(a^2+b^2+c^2-ab-ac-bc\right)}{2.\left(a^2+b^2+c^2-ab-bc-ca\right)}\)( a+b+c=0)
\(P=\frac{5}{2}\left[\left(a^2+b^2+c^2-ab-bc-ca\right)\ne0\right]\)
1. Cho a2 - b2 - c2 =3abc
Tính H = \(\left(1-\frac{a}{b}\right)\left(1-\frac{b}{c}\right)\left(1-\frac{c}{a}\right)\)
2. Cho a - b + c = - 4
Tính B = \(\frac{a^3-b^3+c^3+3abc}{\left(a+b\right)^2+\left(b+c\right)^2+\left(c-a\right)^2}\)
Cho a+b+c\(a^3+b^3+c^3=3abc\) áp dụng tính B=\(\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3}\)
cho a;b;c là các số thực dương.CMR:\(\frac{a+b}{\sqrt{ab+c^2}}+\frac{b+c}{\sqrt{bc+a^2}}+\frac{c+a}{\sqrt{ca+b^2}}\ge4\sqrt{1+\frac{3abc}{\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3}}\)
mình hướng dẫn thôi được không chứ mình đá bóng bị ngã nên giờ bấm giải chi tiết không nổi
thôi mình sẽ giải chi tiết luôn nhé chứ hướng dẫn khó hiểu lắm
đặt cái vế trái là A. Ta có:
\(A=a\left(\frac{1}{\sqrt{ab+c^2}}+\frac{1}{\sqrt{ac+b^2}}\right)+b\left(\frac{1}{\sqrt{ab+c^2}}+\frac{1}{\sqrt{bc+a^2}}\right)+c\left(\frac{1}{\sqrt{ac+b^2}}+\frac{1}{\sqrt{bc+a^2}}\right)\)
\(\Rightarrow A\ge4\left(\frac{a}{\sqrt{ab+c^2}+\sqrt{ac+b^2}}+\frac{b}{\sqrt{ab+c^2}+\sqrt{bc+a^2}}+\frac{c}{\sqrt{ac+b^2}+\sqrt{bc+a^2}}\right)\)
Cho a + b + c = 3. Tính:
M = \(\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
Cho a + b + c = 3. Tính
M = \(\frac{a^3+b^3+c^3-3abc}{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}\)
$Ta$Ta $co$co: $a^3+b^3+c^3-3abc$a3+b3+c3−3abc