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Lời giải:
ĐK....................
a)
\(\frac{a^3-4a^2-a+4}{a^3-7a^3+14a-8}=\frac{(a^3-4a^2)-(a-4)}{(a^3-4a^2)-(3a^2-12a)+(2a-8)}=\frac{a^2(a-4)-(a-4)}{a^2(a-4)-3a(a-4)+2(a-4)}\)
\(=\frac{(a-4)(a^2-1)}{(a-4)(a^2-3a+2)}=\frac{a^2-1}{a^2-3a+2}=\frac{(a-1)(a+1)}{(a-1)(a-2)}=\frac{a+1}{a-2}\) (đpcm)
b)
\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{(x^4+x^3)+(x+1)}{(x^4+x^2)-(x^3+x)+x^2+1}=\frac{x^3(x+1)+(x+1)}{x^2(x^2+1)-x(x^2+1)+(x^2+1)}=\frac{(x+1)(x^3+1)}{(x^2+1)(x^2-x+1)}\)
\(=\frac{(x+1)(x+1)(x^2-x+1)}{(x^2+1)(x^2-x+1)}=\frac{(x+1)^2}{x^2+1}\) (đpcm)
Lời giải:
ĐK....................
a)
\(\frac{a^3-4a^2-a+4}{a^3-7a^3+14a-8}=\frac{(a^3-4a^2)-(a-4)}{(a^3-4a^2)-(3a^2-12a)+(2a-8)}=\frac{a^2(a-4)-(a-4)}{a^2(a-4)-3a(a-4)+2(a-4)}\)
\(=\frac{(a-4)(a^2-1)}{(a-4)(a^2-3a+2)}=\frac{a^2-1}{a^2-3a+2}=\frac{(a-1)(a+1)}{(a-1)(a-2)}=\frac{a+1}{a-2}\) (đpcm)
b)
\(\frac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\frac{(x^4+x^3)+(x+1)}{(x^4+x^2)-(x^3+x)+x^2+1}=\frac{x^3(x+1)+(x+1)}{x^2(x^2+1)-x(x^2+1)+(x^2+1)}=\frac{(x+1)(x^3+1)}{(x^2+1)(x^2-x+1)}\)
\(=\frac{(x+1)(x+1)(x^2-x+1)}{(x^2+1)(x^2-x+1)}=\frac{(x+1)^2}{x^2+1}\) (đpcm)
Đây là câu a/
https://hoc24.vn/hoi-dap/question/693692.html?pos=1903228
Còn câu b thì như sau:
Trước hết, nghi ngờ bạn ghi sai đề ở con này \(\dfrac{1}{a^2+7a+9}\) , số 9 phải là số 12 mới hợp lý. Mình tự sửa lại đề, còn nếu đề đúng như bạn chép thì bạn giữ nguyên nó, phần còn lại rút gọn được còn đâu thì quy đồng giải trâu thôi, chẳng cách nào với đề xấu kiểu ấy cả.
\(B=\dfrac{1}{a\left(a+1\right)}+\dfrac{1}{\left(a+1\right)\left(a+2\right)}+\dfrac{1}{\left(a+2\right)\left(a+3\right)}+\dfrac{1}{\left(a+3\right)\left(a+4\right)}+\dfrac{1}{\left(a+4\right)\left(a+5\right)}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+1}+\dfrac{1}{a+1}-\dfrac{1}{a+2}+\dfrac{1}{a+2}-\dfrac{1}{a+3}+\dfrac{1}{a+3}-\dfrac{1}{a+4}+\dfrac{1}{a+4}-\dfrac{1}{a+5}\)
\(B=\dfrac{1}{a}-\dfrac{1}{a+5}=\dfrac{5}{a\left(a+5\right)}\)
a: \(VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-1\right)\left(a^2-5a+4\right)}\)
\(=\dfrac{\left(a-4\right)\left(a+1\right)}{\left(a-4\right)\left(a-1\right)}=\dfrac{a+1}{a-1}=VP\)
b: \(VT=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x+1\right)^2}{x^2+1}=VP\)
\(\frac{a^3-4a^2-a+a}{a^3-7a^2+14a-8}=\frac{a^3-4a^2}{a^3-4a^2-3a^2+12a+2a-8}\)
\(=\frac{a^2\left(a-4\right)}{a^2\left(a-4\right)-3a\left(a-4\right)+2\left(a-4\right)}=\frac{a^2\left(a-4\right)}{\left(a-4\right)\left(a^2-3a+2\right)}\)
\(=\frac{a^2}{a^2-3a+2}=\frac{a^2}{a\left(a-2\right)-\left(a-2\right)}=\frac{a^2}{\left(a-2\right)\left(a-1\right)}\)
Ủng hộ mik nhé!!!!
P=\(\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a^3-4a^2\right)-\left(a-4\right)}{\left(a^3-8\right)-\left(7a^2-14a\right)}\)
\(=\frac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(=\frac{\left(a-4\right)\left(a^2-1\right)}{\left(a-2\right)\left(\left(a^2-4a\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a\left(a-4\right)-\left(a-4\right)\right)}\)
\(=\frac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a-4\right)\left(a-1\right)}\)
\(=\frac{a+1}{a-2}\)
Chúc bạn học giỏi, k cho mình nhé!!!
Câu 1:
Ta có: \(\left(\dfrac{a+b}{2}\right)^2\ge ab\)
\(\Leftrightarrow\dfrac{\left(a+b\right)^2}{2^2}-ab\ge0\)
\(\Leftrightarrow\dfrac{a^2+2ab+b^2-4ab}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab+b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\left(\dfrac{a+b}{2}\right)^2\ge ab\) (1)
Ta có: \(\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\)
\(\Leftrightarrow\dfrac{a^2+b^2}{2}-\dfrac{\left(a+b\right)^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{2a^2-2b^2-a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{a^2-2ab-b^2}{4}\ge0\)
\(\Leftrightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\)
Vì \(\left(a-b\right)^2\ge0\forall a,b\)
\(\Rightarrow\dfrac{\left(a-b\right)^2}{4}\ge0\forall a,b\)
\(\Rightarrow\dfrac{a^2+b^2}{2}\ge\left(\dfrac{a+b}{2}\right)^2\) (2)
Từ (1) và (2) \(\Rightarrow ab\le\left(\dfrac{a+b}{2}\right)^2\le\dfrac{a^2+b^2}{2}\)
5 , a3+b3+c3\(\ge\) 3abc
\(\Leftrightarrow\) a3+3a2b+3ab2+b3+c3-3a2b-3ab2-3abc\(\ge\) 0
\(\Leftrightarrow\) (a+b)3+c3-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+2ab+b2-ac-bc+c2)-3ab(a+b+c) \(\ge0\)
\(\Leftrightarrow\) (a+b+c)(a2+b2+c2-ab-bc-ca)\(\ge0\) (1)
ta co : a,b,c>0 \(\Rightarrow\)a+b+c>0 (2)
(a-b)2+(b-c)2+(c-a)2\(\ge0\)
<=> 2a2+2b2+2c2-2ac-2cb-2ab\(\ge0\)
<=>a2+b2+c2-ab-bc-ac\(\ge\) 0 (3)
Từ (1)(2)(3)=> pt luôn đúng
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
a: \(=\dfrac{3x\left(x-y\right)^2\cdot\left(x-1\right)}{3x\left(x-1\right)\cdot\left(x-y\right)^2\cdot2\cdot\left(x-y\right)}=\dfrac{1}{2\left(x-y\right)}\)
b: =(x+1)^2/(x+1)=x+1
c: \(=\dfrac{a\left(a^2-4a+4\right)}{\left(a-2\right)\left(a+2\right)}=\dfrac{a\left(a-2\right)^2}{\left(a-2\right)\left(a+2\right)}=\dfrac{a\left(a-2\right)}{a+2}\)
d: \(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
Có \(\text{VT }=\) \(\dfrac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}\)
\(\Rightarrow VT=\dfrac{a^2\left(a-4\right)-\left(a-4\right)}{\left(a-2\right)\left(a^2+2a+4\right)-7a\left(a-2\right)}\)
\(\Rightarrow VT=\dfrac{\left(a-4\right)\left(a-1\right)\left(a+1\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow VT=\dfrac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}\)
\(\Rightarrow\dfrac{a+1}{a-2}\)
\(\Rightarrow VT=VP\)
\(\Rightarrowđpcm\)