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D=\(\dfrac{abc+a+b+c-1-ab-bc-ca}{a^2b+1-a^2-b}\)
\(=\dfrac{\left(abc-bc\right)-\left(ca-c\right)-\left(ab-b\right)+\left(a-1\right)}{\left(a^2b-a^2\right)+\left(1-b\right)}\)
\(=\dfrac{bc\left(a-1\right)-c\left(a-1\right)-b\left(a-1\right)+\left(a-1\right)}{a^2\left(b-1\right)+\left(1-b\right)}\)
\(=\dfrac{\left(a-1\right)\left(bc-c-b+1\right)}{a^2\left(b-1\right)-\left(b-1\right)}=\dfrac{\left(a-1\right)\left[\left(bc-c\right)-\left(b-1\right)\right]}{\left(b-1\right)\left(a^2-1\right)}\)
\(=\dfrac{\left(a-1\right)\left[c\left(b-1\right)-\left(b-1\right)\right]}{\left(b-1\right)\left(a-1\right)\left(a+1\right)}=\dfrac{\left(a-1\right)\left(b-1\right)\left(c-1\right)}{\left(b-1\right)\left(a-1\right)\left(a+1\right)}\)
\(=\dfrac{c-1}{a+1}\)
ai có thể giảng cho mình dạng toán tìm số tự nhiên thỏa mãn đièu kiện chia hết ko
hãy nêu ra cách giải cụ thể cho câu sau 3a-11 chia hết cho a+2 tìm a
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)
=\(2\left(a+b+c\right)+\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}-\frac{ab}{c}-\frac{bc}{a}-\frac{ca}{b}=2\left(a+b+c\right)\)
\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)
=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)
2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)
\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+bc\right)+\left(a+b+c\right)ac-abc\)
\(=\left(ab+b^2+bc\right)\left(a+c\right)+\left(a+c\right)ac+abc-abc\)
\(=\left(a+c\right)\left(ab+b^2+bc+ac\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
nhân cả vế với abc ta có điều cần chứng minh
\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\ge\dfrac{ab+bc+ac}{2}\)
VT\(\ge\)\(\dfrac{\left(bc+ac+ab\right)^2}{2\left(ab+bc+ac\right)}=\dfrac{bc+ac+ab}{2}\)
=>(đpcm)
mấu chốt nằm ở đoạn chứng minh\(\dfrac{\left(bc\right)^2}{a\left(b+c\right)}+\dfrac{\left(ac\right)^2}{b\left(a+c\right)}+\dfrac{\left(ab\right)^2}{c\left(a+b\right)}\)
chỉ cần chứng minh được \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\)sau đó áp dụng để chứng minh cái kia thôi cái này bạn thử tự chứng minh nhé
\(P=\dfrac{abc+a+b+c-\left(ab+bc+ca+1\right)}{a^2b+1-\left(a^2+b\right)}\)
\(=\dfrac{abc+a+b+c-ab-bc-ca-1}{a^2b+1-a^2-b}\)
\(=\dfrac{\left(a-1\right)+\left(abc-bc\right)-\left(ab-b\right)-\left(ca-c\right)}{\left(a^2b-a^2\right)-\left(b-1\right)}\)
\(=\dfrac{\left(a-1\right)+bc\left(a-1\right)-b\left(a-1\right)-c\left(a-1\right)}{a^2\left(b-1\right)-\left(b-1\right)}\)
\(=\dfrac{\left(a-1\right)\left(1+bc-b-c\right)}{\left(b-1\right)\left(a^2-1\right)}\)
\(=\dfrac{\left(a-1\right)\left[b\left(c-1\right)-\left(c-1\right)\right]}{\left(b-1\right)\left(a-1\right)\left(a+1\right)}\)
\(=\dfrac{\left(b-1\right)\left(c-1\right)}{\left(b-1\right)\left(a+1\right)}=\dfrac{c-1}{a+1}\)
\(\text{#}Toru\)