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a) A=2m(n-p)-(n-p)(m+p)
=(2m-m-p)(n-q)
=(m-p)(n-q)
thay m=18,3; n=24,6; p=10,6; q=-31,7 vào biểu thức trên ta có
A=(18,3-10,6)(24,6+31,7)
=7,7.56,3
=433,51
b) B=(a-b)(b+c)+b(b-a)
=(a-b)(b+c)-b(a-b)
=(a-b)(b+c-b)
=(a-b)c
thay a=0,68;b=0,26;c=1,5 vào biểu thức ta có
B=(2-1,007)(-0,006)
=0.993.(-0,006)
=-0,005958
1.
Ta có: \(a^4+b^4\ge\frac{1}{2}\left(a^2+b^2\right)\left(a^2+b^2\right)\ge ab\left(a^2+b^2\right)\)
\(\Rightarrow VT\le\frac{a}{a+bc\left(b^2+c^2\right)}+\frac{b}{b+ca\left(c^2+a^2\right)}+\frac{c}{c+ab\left(a^2+b^2\right)}\)
\(\Rightarrow VT\le\frac{a^2}{a^2+abc\left(b^2+c^2\right)}+\frac{b^2}{b^2+abc\left(a^2+c^2\right)}+\frac{c^2}{c^2+abc\left(a^2+b^2\right)}\)
\(\Rightarrow VT\le\frac{a^2}{a^2+b^2+c^2}+\frac{b^2}{a^2+b^2+c^2}+\frac{c^2}{a^2+b^2+c^2}=1\)
Dấu "=" xảy ra khi \(a=b=c=1\)
\(a.a\left(b+c\right)+3b+3c=a\left(b+c\right)+3\left(b+c\right)=\left(b+c\right)\left(a+3\right)\)
\(b.a\left(c-d\right)+c-d=\left(c-d\right)\left(a+1\right)\)
\(c.b\left(a-c\right)+5a-5c=b\left(a-c\right)+5\left(a-c\right)=\left(a-c\right)\left(b+5\right)\)
\(d.a\left(m-n\right)+m-n=\left(m-n\right)\left(a+1\right)\)
\(e.mx+my+5x+5y=m\left(x+y\right)+5\left(x+y\right)=\left(x+y\right)\left(m+5\right)\)
\(f.ma+mb-a-b=m\left(a+b\right)-\left(a+b\right)=\left(a+b\right)\left(m-1\right)\)
\(g.4x+by+4y+bx=4x+bx+by+4y=x\left(b+4\right)+y\left(b+4\right)=\left(b+4\right)\left(x+y\right)\)
\(h.1-ax-x+a=\left(a+1\right)-x\left(a+1\right)=\left(a+1\right)\left(1-x\right)\)
\(k.x^{m+2}-x^m=x^m\left(x^2-1\right)=x^m\left(x-1\right)\left(x+1\right)\)
\(m.\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)=\left(b-a\right)^2-\left(b-a\right)\left(a+b\right)=\left(b-a\right)\left(b-a-a-b\right)=-2a\left(b-a\right)\)
\(n.a\left(a-b\right)\left(a+b\right)-\left(a^2-2ab+b^2\right)=a\left(a-b\right)\left(a+b\right)-\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab-a+b\right)\)
a) Áp dụng bất đẳng thức Schur với \(r=1\)
\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)
\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)
\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
b) Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)
\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )
Dấu " = " xảy ra khi \(a=b=c\)
c) Ta có \(abc=ab+bc+ca\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)
Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)
\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )
Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0
\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)
\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\le\dfrac{3}{16}\)
\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )
1, a +b +c = 0 => a + b = -c ; a +c = -b ; b+c = -a
thay vào M ta có
M = a . -c . -b = abc (1)
Thay tương tự vào N , P ta cũng đc N =abc (2)
P =abc( 3)
Từ 1 2 và 3 => ĐPCM
2,
a + b +c = 2P
=> b + c = 2P -a
=> ( b + c)^2 = ( 2P -a)^2
=> b^2 + 2bc+ c^2 = 4p^2 - 4pa + a^2
=> 2bc+ b^2 + c^2 -a^ 2 = 4p^2 - 4pa
=> 2bc + b^2 + c^2 -a ^ 2 = 4p(p-a)=> ĐPCM
a) \(\cdot\left(m+n\right)^2-\left(m-n\right)^2+\left(m+n\right)\left(m-n\right)\)
\(=\left(m+n+m-n\right)\left(m+n-m+n\right)+\left(m+n\right)\left(m-n\right)\)
\(=\left(2m\cdot2n\right)+m^2-n^2\)
\(=4mn+m^2-n^2\)
b) \(\left(a+b\right)^2-\left(a-b\right)^2-2a^3\)
\(=\left(a+b+a-b\right)\left(a+b-a+b\right)-2a^3\)
\(=2ab-2a^3\)
c) \(\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=\left(2x+1\right)^2+2\left(2x+1\right)\left(2x-1\right)+\left(2x-1\right)^2\)
\(=\left(2x+1+2x-1\right)^2\)
\(=\left(4x\right)^2=16x^2\)
d) \(\left(a+b+c\right)^2-2\left(a+b+c\right)\left(b+c\right)+\left(b+c\right)^2\)
\(=\left(a+b+c-b-c\right)^2=a^2\)
xin lỗi mk ghi sai đề ở bài :d) (a+b+c)^2-2(a+b+c)(b+c)+(b+c)^2