thu gọn
a) P=13(4x-1)+20x-190
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KT
10 tháng 7 2018
Bài 1:
\(A=x^2-6x+13=\left(x-3\right)^2+4\ge4\)
Vậy \(Min\)\(A=4\)\(\Leftrightarrow\)\(x=3\)
\(B=2x^2+8x=2\left(x^2+4x+4\right)-8=2\left(x+2\right)^2-8\ge-8\)
Vậy \(Min\)\(B=-8\)\(\Leftrightarrow\)\(x=-2\)
\(C=4x^2+20x=\left(2x+5\right)^2-25\ge-25\)
Vậy \(Min\)\(C=-25\)\(\Leftrightarrow\)\(x=-\frac{5}{2}\)
Bài 3:
a) \(x^2+12x+39=\left(x+6\right)^2+3>0\)
b) \(4x^2+4x+3=\left(2x+1\right)^2+2>0\)
9 tháng 12 2018
\(\frac{4x^4-20x^3+13x^2+30x+9}{\left(4x^2-1\right)^2}\)
\(=\frac{4x^3\left(x-3\right)-8x^2\left(x-3\right)-11x\left(x-3\right)-3\left(x-3\right)}{\left(4x^2-1\right)^2}\)
\(=\frac{\left(x-3\right)\left(4x^3-8x^2-11x-3\right)}{\left(4x^2-1\right)^2}\)
\(=\frac{\left(x-3\right)\left[4x^2\left(x-3\right)+4x\left(x-3\right)+\left(x-3\right)\right]}{\left[\left(2x-1\right)\left(2x+1\right)\right]^2}\)
\(=\frac{\left(x-3\right)^2\left(4x^2+4x+1\right)}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2\left(2x+1\right)^2}{\left(2x-1\right)^2\left(2x+1\right)^2}=\frac{\left(x-3\right)^2}{\left(2x-1\right)^2}\)
A
tìm x biết \(|x+1|+|x+\frac{1}{3}|+|x+\frac{1}{6}|+|x+\frac{1}{10}|+...+|x+\frac{1}{190}|=20x\) =20x
2
24 tháng 11 2019
Ta có \(\left|x+1\right|\ge0;\left|x+\frac{1}{3}\right|\ge0;...;\)\(\left|x+\frac{1}{190}\right|\ge0\) \(\forall x\)
=> \(\left|x+1\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{190}\right|\ge0\) \(\forall x\)
=> \(20x\ge0\Rightarrow x\ge0\)
Với \(x\ge0\) => \(x+1>0,x+\frac{1}{3}>0,x+\frac{1}{6}>0,...,x+\frac{1}{190}>0\)
=> \(\left|x+1\right|=x+1,\left|x+\frac{1}{3}\right|=x+\frac{1}{3},\left|x+\frac{1}{6}\right|=x+\frac{1}{6},...,\left|x+\frac{1}{190}\right|=x+\frac{1}{190}\)
=> \(x+1+x+\frac{1}{3}+x+\frac{1}{6}+...+x+\frac{1}{190}=20x\)
=> \(19x+\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)=20x\)
=> \(x=\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{190}\right)\)
Gọi \(A=1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{190}\)
=> \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{380}\)
=> \(\frac{1}{2}A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{19.20}\)
=> \(\frac{1}{2}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\)
=> \(\frac{1}{2}A=1-\frac{1}{20}\)
=> \(A=\frac{19}{10}\)
Thay vào ta có
=> \(x=-\frac{19}{10}\)
21 tháng 1 2024
\(\dfrac{20x^2+120x+180}{\left(3x+5\right)^2-4x^2}+\dfrac{5x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{3\left(x^2+8x+15\right)}\)
\(=\dfrac{20\left(x^2+6x+9\right)}{\left(3x+5+2x\right)\left(3x+5-2x\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{20\left(x+3\right)^2}{5\left(x+1\right)\cdot\left(x+5\right)}+\dfrac{5\left(x-5\right)\left(x+5\right)}{5\left(x+1\right)\left(x-5\right)}-\dfrac{\left(x+3\right)\cdot3\left(x+1\right)}{3\left(x+3\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(x+1\right)\left(x+5\right)}+\dfrac{x+5}{x+1}-\dfrac{x+1}{x+5}\)
\(=\dfrac{4\left(x+3\right)^2+\left(x+5\right)^2-\left(x+1\right)^2}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4x^2+24x+36+x^2+10x+25-x^2-2x-1}{\left(x+1\right)\cdot\left(x+5\right)}\)
\(=\dfrac{4x^2+32x+60}{\left(x+1\right)\left(x+5\right)}=\dfrac{4\left(x^2+8x+15\right)}{\left(x+1\right)\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)\cdot\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\dfrac{4x+12}{x+1}\)
DT
0
15 tháng 4 2020
Câu a và b mình ko viết đề nhé bạn!
a)=\(\left(\frac{3}{5}.\frac{5}{7}\right).\left(x^2.x^4\right).\left(y^2.y^5\right)\)
=\(\frac{3}{7}x^6y^7\)
Hệ số:\(\frac{3}{7}\)
Phần biến:\(x^6y^7\)
b)=\(\left(-20\right).\frac{1}{5}.\left(x^4.x\right).\left(y^2.y\right)\)
=\(-4x^5y^3\)
Hệ số:\(-4\)
Phần biến:\(x^5y^3\)
Nhớ tick cho mình nha!
a) P=13(4x-1)+20x-190
=52x-13+20x-190
=(52x+20x)+(-13-190)
=72x-203
\(P=13\left(4x-1\right)+20x-190\)
\(=52x-13+20x-190=72x-203\)