thu gọn các đơn thức sau
a)ab.4/3a^2b^4.7abc
b)a^3b^3.a^2b^2c
c)2/3a^3b.(-1/2ab).a^2b
d)-2 1/3a^3c^21/7ac^2 6abc
e)(-1,5ab^2)1/4bca^2b
thu gọn các đa thức sau:
a,2a^3.(-1/2ab).a^2b
b,-2/1/3a^3c^2.1/7ac^2.6abc
c,2ab.4/3a^2b^4.7abc
d,2y.3y^2.d^2y^2
e,(-2/1/3.cd).(1/1/4c^2d).(-5/6cd)^2
g,(1/2a.1/4a^2.1/8^3)^2.2b.4b^2-8b^3
Tính các tích sau:
a) 5xy.(-2bx2y)
b) \(\left(\frac{-4}{5}ab^2c\right)\left(-20a^4bx\right)\)
c) \(^{2^3abc.\frac{1}{4}a^2bc^3}\)
d) \(a^3b^3a^2b^2c\)
e) \(2ab.\frac{4}{3}a^2b^4.7abc\)
f) \(\left(-1,5ab^2\right)\frac{1}{4}bca^2b\)
***Lm giúp mk vs mk cần gấp<3
a) \(5xy\cdot\left(-2bx^2y\right)=-10b\left(x\cdot x^2\right)\left(y\cdot y\right)=-10bx^3y^2\)
b) \(\left(-\frac{4}{5}ab^2c\right)\left(-20a^4bx\right)=\left[\left(-\frac{4}{5}\right)\cdot\left(-20\right)\right]\left(a\cdot a^4\right)\left(b^2\cdot b\right)cx\)
\(=16a^5b^3cx\)
c) \(2^3abc\cdot\frac{1}{4}a^2bc^3=8abc\cdot\frac{1}{4}a^2bc^3=2\left(a\cdot a^2\right)\left(b\cdot b\right)\left(c\cdot c^3\right)=2a^3b^2c^4\)
d) \(a^3b^3a^2b^2c=\left(a^3\cdot a^2\right)\left(b^3\cdot b^2\right)c=a^5b^5c\)
e) \(2ab\cdot\frac{4}{3}a^2b^4\cdot7abc=\left(2\cdot\frac{4}{3}\cdot7\right)\left(a\cdot a^2\cdot a\right)\left(b\cdot b^4\cdot b\right)c=\frac{56}{3}a^4b^6c\)
f) \(\left(-1,5ab^2\right)\cdot\frac{1}{4}bca^2b=\left(-1,5\cdot\frac{1}{4}\right)\left(a\cdot a^2\right)\left(b^2\cdot b\cdot b\right)c=-\frac{3}{8}a^3b^4c\)
Tính các tích sau:
a) 5xy.(-2bx2y)
b) \(\left(\frac{-4}{5}ab^2c\right)\left(-20a^4bx\right)\)
c) \(^{2^3abc.\frac{1}{4}a^2bc^3}\)
d) \(a^3b^3a^2b^2c\)
e) \(2ab.\frac{4}{3}a^2b^4.7abc\)
f) \(\left(-1,5ab^2\right)\frac{1}{4}bca^2b\)
***Lm giúp mk vs mk cần gấp<3
a) \(5xy.\left(-2bx^2y\right)\)
\(=\left[5.\left(-2\right)\right]\left(x.x^2\right)\left(y.y\right).b\)
\(=-10x^3y^2b\)
b) \(\left(-\frac{4}{5}ab^2c\right)\left(-20a^4bx\right)\)
\(=\left[\left(-\frac{4}{5}\right)\left(-20\right)\right]\left(a.a^{4\:}\right)\left(b^2b\right).c.x\)
\(=16a^5b^3cx\)
c) \(2^3abc.\frac{1}{4}a^2bc^3\)
\(=\left(2^3.\frac{1}{4}\right)\left(aa^{2\:}\right)\left(bb\right)\left(cc^3\right)\)
\(=2a^3b^2c^4\)
d) \(a^3b^3a^2b^2c\)
\(=\left(a^3a^2\right)\left(b^3b^2\right)c\)
\(=a^5b^5c\)
e) \(2ab.\frac{4}{3}a^2b^47abc\)
\(=\left(2.\frac{4}{3}.7\right)\left(aa^{2\: }a\right)\left(bb^4b\right)c\)
\(=\frac{56}{3}a^4b^6c\)
f) \(\left(-1,5ab^2\right)\frac{1}{4}bca^2b\)
\(=\left(-1,5.\frac{1}{4}\right)\left(aa^{2\:}\right)\left(b^2bb\right)\)
\(=-\frac{3}{8}a^3b^4\)
1. Cho \(a,b,c>0\) và \(ab+bc+ca=abc\). Chứng minh rằng:
\(\dfrac{1}{a+3b+2c}+\dfrac{1}{b+3c+2a}+\dfrac{1}{c+3a+2b}\le\dfrac{1}{6}\)
2. Cho \(a,b\ge0\) và \(a+b=2\) Tìm Max
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+20ab\)
Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)
Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)
CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)
\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)
Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)
Dấu = xảy ra khi a=b=c=3
Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)
\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)
\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)
\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)
Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)
\(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)
\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)
Vậy...
2,
\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)
\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)
\(=9a^2b^2-2ab+48\)
Đặt \(ab=x\Rightarrow0\le x\le1\)
\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)
\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)
cmr: (a+2b-3c)^3+(b+2c-3a)^3+(c+2a-3b)^3=3.(a+2b-3c).(b+2c-3a).(c+2a-3b)
Cho a+b+c = 1 và 3a+2b>c, 3b+2c>a, 3c+2a>b. Chứng minh: 1/(3a+2b-c) + 1/(3b+2c-a) + 1/(3c+2a-b) >hoặc = 9/4
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh:
1) \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2) \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3) \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4) \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=bk;c=dk\)
1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)
\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)
Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)
2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)
\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)
3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)
Do đó: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3c+5d}{3c-5d}\)
4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)
\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)
Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)
CMR: Với mọi a;b;c>0
\(\frac{2b+3c}{a+2b+3c}+\frac{2c+3a}{b+2c+3a}+\frac{2a+3b}{c+2a+3b}\ge\frac{5}{2}\)
Với a,b,c thuộc R thỏa mãn :
CMR : (a+2b)(b+2c)(c+2a)=1
Lời giải:
Đặt ⎧⎪⎨⎪⎩3a+b−c=x3b+c−a=y3c+a−b=z{3a+b−c=x3b+c−a=y3c+a−b=z
Khi đó, điều kiện đb tương đương với:
(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24
⇔3(2a+4b)(2b+4c)(2c+4a)=24⇔3(2a+4b)(2b+4c)(2c+4a)=24
⇔(a+2b)(b+2c)(c+2a)=1⇔(a+2b)(b+2c)(c+2a)=1
Do đó ta có đpcm
Lời giải:
Đặt ⎧⎪⎨⎪⎩3a+b−c=x3b+c−a=y3c+a−b=z{3a+b−c=x3b+c−a=y3c+a−b=z
Khi đó, điều kiện đb tương đương với:
(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24(x+y+z)3=24+x3+y3+z3⇔3(x+y)(y+z)(x+z)=24
⇔3(2a+4b)(2b+4c)(2c+4a)=24⇔3(2a+4b)(2b+4c)(2c+4a)=24
⇔(a+2b)(b+2c)(c+2a)=1⇔(a+2b)(b+2c)(c+2a)=1
Do đó ta có đpcm