mk ms 2k8

( 2 + 2/3 - 1/4 ) * 1/5

= ( 6/ 3 + 2/3 - 1/4 ) * 1/5

= ( 8/3 - 1/4 ) * 1/5 

= ( 32/12 - 3/ 12 ) * 1/5

= 29 / 12 * 1/5 

=29/60 

5/ Tưỡng dễ ăn = sos + bđt phụ ai ngờ....hic...

\(BĐT\Leftrightarrow\Sigma_{cyc}\left(\frac{a^2+b^2+c^2}{a+b+c}-\frac{a^2+b^2}{a+b}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\left(\frac{\left(a^2+b^2+c^2\right)\left(a+b\right)-\left(a^2+b^2\right)\left(a+b+c\right)}{\left(a+b+c\right)\left(a+b\right)}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)-bc\left(b-c\right)}{\left(a+b+c\right)\left(a+b\right)}\ge0\)\(\Leftrightarrow\Sigma_{cyc}\left(\frac{ca\left(c-a\right)}{\left(a+b+c\right)\left(a+b\right)}-\frac{ca\left(c-a\right)}{\left(a+b+c\right)\left(b+c\right)}\right)\ge0\)

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)^2}{\left(a+b+c\right)}\ge0\left(\text{đúng}\right)\)

Ai ngờ nổi khi không dùng BĐT phụ lại dễ hơn cái kia chứ -_-

Ây za,nhầm dòng cuối cùng xíu ạ:

\(\Leftrightarrow\Sigma_{cyc}\frac{ca\left(c-a\right)^2}{\left(a+b+c\right)\left(a+b\right)\left(b+c\right)}\ge0\left(\text{đúng}\right)\) -_- đánh thiếu một chút lại ra nông nỗi -_-

Bài 1:

Xét các hiệu sau:

\(M=x^3+y^3+z^3-(x^2+y^2+z^2)=x^2(x-1)+y^2(y-1)+z^2(z-1)\)

\(N=x^4+y^4+z^4-(x^3+y^3+z^3)=x^3(x-1)+y^3(y-1)+z^3(z-1)\)

Lấy $N-M$:

\( N-M=\sum x^2(x-1)(x-1)=\sum x^2(x-1)^2\geq 0\)

\(\Leftrightarrow \sum x^4-2\sum x^3+\sum x^2\geq 0\)

\(\Rightarrow \sum x^4\geq 2\sum x^3-\sum x^2(*)\)

\(P=x^5+y^5+z^5-(x^4+y^4+z^4)=x^4(x-1)+y^4(y-1)+z^4(z-1)\)

Lấy $P-M$

\(P-M=\sum x^2(x-1)(x^2-1)=\sum x^2(x-1)^2(x+1)\geq 0, \forall x,y,z>-1\)

\(\Leftrightarrow \sum x^5-\sum x^4-\sum x^3+\sum x^2\geq 0\)

\(\Leftrightarrow \sum x^5\geq \sum x^4+\sum x^3-\sum x^2\). Kết hợp với (*) và điều kiện ban đầu suy ra:

\(\sum x^5\geq 2\sum x^3-\sum x^2+\sum x^3-\sum x^2=3\sum x^3-2\sum x^2\geq \sum x^2\)

Ta có

△\(=b^2-4ac=\left[-\left(m+1\right)\right]^2-4.1.\left(m-5\right)=m^2+2m+1-4m+20=m^2-2m+21>0\)Vậy phương trình luôn có 2 nghiệm \(x_1,x_2\) phân biệt

Theo định lí Vi-ét ta có

\(\left\{{}\begin{matrix}x_1+x_2=\frac{-b}{a}=\frac{m+1}{1}=m+1\\x_1x_2=\frac{c}{a}=\frac{m-5}{1}=m-5\end{matrix}\right.\)

Ta lại có \(x_1^3-x_2^3=32\Leftrightarrow\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)=32\Leftrightarrow4.\left(x_1^2+2x_1x_2+x_2^2-x_1x_2\right)=32\Leftrightarrow\left(x_1+x_2\right)^2-x_1x_2=8\Leftrightarrow\left(m+1\right)^2-\left(m-5\right)=8\Leftrightarrow m^2+2m+1-m+5-8=0\Leftrightarrow m^2+m-2=0\Leftrightarrow\left(m-1\right)\left(m+2\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}m=1\\m=-2\end{matrix}\right.\)

Vậy m=1 hoặc m=-2 thì phương trình có 2 nghiệm \(x_1,x_2\) thỏa mãn \(\left\{{}\begin{matrix}x_1-x_2=4\\x_1^3-x_2^3=32\end{matrix}\right.\)

a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)

\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)

\(=4-2\sqrt{2}\)

b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)

\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)

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

a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)

\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)

\(P=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}\)

\(P=2\)

b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)

\(N=\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)

\(N=1^2-\left(\sqrt{5}\right)^2\)

\(N=-4\)

c) \(Q=\left(\dfrac{5+2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)

\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)

\(Q=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)

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

\(Q=1\)

Cả 4 ý đều sai bạn nhé! 

Cả bốn câu sai đúng như mình suy rằng cả bốn phép tính đều sai còn các bạn khác có như đáp án của mình và khánh lưa ko nhớ nhắn cho mình nhé hi hi😀😂

cả 4 ý đều sai nhé

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144