Tìm GTLN của
a) A = \(-x^2-y^2-2x+y+1\)
b) B= \(\dfrac{-1-2x+x^2}{x^2}\)
c) C = \(\dfrac{-2x}{x^2+1}\)
tìm điều kiện bài toán:
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt{2x-1}}{x^2-3x+2}\)
b) \(y=\dfrac{1}{x^2-1}-\sqrt{7-2x}\)
c) \(y=\dfrac{2}{x}+\dfrac{3}{4-2x+x^2}\)
d) \(y=\sqrt{25-x^2}-2\sqrt{x}+3\)
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
a) \(x\ge\dfrac{1}{2};x\ne1;x\ne2\)
b) \(x\le\dfrac{7}{2};x\ne\pm1\)
c) \(x\ne0\)
d) \(0\le x\le5\)
1, Cho x; y; z ≠0 và \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\)+ \(\dfrac{1}{z}\)=\(\dfrac{2}{2x+y+2z}\). Cmr: (2x+y)(y+2z)(z+x)= 0
2, Cho \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=1\). Cmr: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
Gấp ạ, ai giúp mình với!!!!
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
3. Tìm GTLN, GTNN:
a) \(y=2\sin^2x+3\sin x\cos x-2\cos^2x+5\)
b) \(y=\dfrac{3\sin x-\cos x+1}{\sin x-2\cos x+4}\)
c) \(y=\dfrac{2\left(x^2+6xy\right)}{1+2xy+y^2}\) biết x, y thay đổi thỏa mãn \(x^2+y^2=1\)
a.
\(y=\dfrac{3}{2}sin2x-2\left(cos^2x-sin^2x\right)+5=\dfrac{3}{2}sin2x-2cos2x+5\)
\(=\dfrac{5}{2}\left(\dfrac{3}{5}sin2x-\dfrac{4}{5}cos2x\right)+5=\dfrac{5}{2}sin\left(2x-a\right)+5\) (với \(cosa=\dfrac{3}{5}\))
\(\Rightarrow-\dfrac{5}{2}+5\le y\le\dfrac{5}{2}+5\)
b.
\(\Leftrightarrow y.sinx-2y.cosx+4y=3sinx-cosx+1\)
\(\Leftrightarrow\left(y-3\right)sinx+\left(1-2y\right)cosx=1-4y\)
Theo điều kiện có nghiệm của pt lượng giác bậc nhất:
\(\left(y-3\right)^2+\left(1-2y\right)^2\ge\left(1-4y\right)^2\)
\(\Leftrightarrow11y^2+2y-9\le0\)
\(\Leftrightarrow-1\le y\le\dfrac{9}{11}\)
c.
Do \(x^2+y^2=1\Rightarrow\) đặt \(\left\{{}\begin{matrix}x=sina\\y=cosa\end{matrix}\right.\)
\(\Rightarrow y=\dfrac{2\left(sin^2a+6sina.cosa\right)}{1+2sina.cosa+cos^2a}=\dfrac{1-cos2a+6sin2a}{1+sin2a+\dfrac{1+cos2a}{2}}=\dfrac{2-2cos2a+12sin2a}{3+2sin2a+cos2a}\)
\(\Leftrightarrow3y+2y.sin2a+y.cos2a=2-2cos2a+12sin2a\)
\(\Leftrightarrow\left(2y-12\right)sin2a+\left(y+2\right)cos2a=2-3y\)
Theo điều kiện có nghiệm của pt bậc nhất theo sin2a, cos2a:
\(\left(2y-12\right)^2+\left(y+2\right)^2\ge\left(2-3y\right)^2\)
\(\Leftrightarrow y^2+8y-36\le0\)
\(\Rightarrow-4-2\sqrt{13}\le y\le-4+2\sqrt{13}\)
Bài 1
a) Tìm GTNN của A = \(\dfrac{2x^2-16x+43}{x^2-8x+22}\)
b) Tìm GTLN của B = \(\dfrac{3x^2+9x+17}{3x^2+9x+7}\)
Bài 2: Tìm x để phân thức có giá trị nguyên
a) \(\dfrac{-6}{3x-2}\) b) \(\dfrac{2x+3}{x-5}\) c) \(\dfrac{x^3-x^2+2}{x-1}\) d) \(\dfrac{2x^3+x^2+2x+2}{2x+1}\) e) \(\dfrac{3x^3-7x^2+11x-1}{3x-1}\)
Bài 3: Cho biểu thức
A= \(\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x^2+10x}\)
a) Rút gọn b) Tìm x để A = 1; A = 3
Bài 4: Cho x + y + z = 0, tính
P= \(\dfrac{x^2}{y^2+z^2-x^2}+\dfrac{y^2}{z^2+x^2-y^2}+\dfrac{z^2}{x^2+y^2-z^2}\)
a) \(A = \frac{2x^2 - 16x+43}{x^2-8x+22}\) = \(\frac{2(x^2-8x+22)-1}{x^2-8x+22}\) = \(2 - \frac{1}{x^2-8x+22}\)
Ta có : \(x^2-8x+22 \) = \(x^2-8x+16+6 = ( x-4)^2 +6 \)
Vì \((x-4)^2 \ge 0 \) với \( \forall x\in R\) Nên \(( x-4)^2 +6 \ge 6 \)
\(\Rightarrow \) \(x^2-8x+22 \) \( \ge 6\)\(\Rightarrow \) \(\frac{1}{x^2-8x+22} \) \(\le \frac{1}{6}\) \(\Rightarrow \) - \(\frac{1}{x^2-8x+22} \) \(\ge - \frac{1}{6}\)
\(\Rightarrow \) A = \(2 - \frac{1}{x^2-8x+22}\) \( \ge 2-\frac{1}{6}\) = \(\frac{11}{6}\) Dấu "=" xảy ra khi và chỉ khi x=4
Vậy GTNN của A = \(\frac{11}{6}\) khi và chỉ khi x=4
Quy đồng mẫu thức của các phân thức
1. \(\dfrac{x-y}{2x^2-4xy+2y^2};\dfrac{x+y}{2x^2+4xy+2y^2};\dfrac{1}{y^2-x^2}\)
2. \(\dfrac{1}{x^2+8x+15};\dfrac{1}{x^2+6x+9}\)
3. \(\dfrac{1}{\left(a-b\right)\left(b-c\right)};\dfrac{1}{\left(c-b\right)\left(c-a\right)};\dfrac{1}{\left(b-a\right)\left(a-c\right)}\)
1: \(MTC=2\left(x-y\right)\left(x+y\right)\)
\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)
\(\dfrac{x+y}{2x^2+4xy+2y^2}\)
\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)
\(=\dfrac{x+y}{2\left(x+y\right)^2}=\dfrac{1}{2\left(x+y\right)}=\dfrac{x-y}{2\left(x+y\right)\left(x-y\right)}\)
\(\dfrac{1}{x^2-y^2}=\dfrac{2}{2\left(x^2-y^2\right)}=\dfrac{2}{2\left(x-y\right)\left(x+y\right)}\)
2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)
3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
Cho a, b, c > 0 thỏa mãn a + b + c = 3. Tìm GTLN của
\(P=\dfrac{x}{\left(2x+y+z\right)^2}+\dfrac{y}{\left(2y+x+z\right)^2}+\dfrac{z}{\left(2z+y+x\right)^2}\)
Chắc đề là \(x+y+z=3\)
Ta có:
\(\left(2x+y+z\right)^2=\left(x+y+x+z\right)^2\ge4\left(x+y\right)\left(x+z\right)\)
\(\Rightarrow P\le\dfrac{x}{4\left(x+y\right)\left(x+z\right)}+\dfrac{y}{4\left(x+y\right)\left(y+z\right)}+\dfrac{z}{4\left(x+z\right)\left(y+z\right)}\)
\(\Rightarrow P\le\dfrac{x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)}{4\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\dfrac{xy+yz+zx}{2\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Mặt khác:
\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(xy+yz+zx\right)\left(x+y+z\right)-xyz\)
\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)
\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)
\(=\dfrac{8}{9}\left(x+y+z\right)\left(zy+yz+zx\right)=\dfrac{8}{3}\left(xy+yz+zx\right)\)
\(\Rightarrow P\le\dfrac{xy+yz+zx}{2.\dfrac{8}{3}\left(xy+yz+zx\right)}=\dfrac{3}{16}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
1. Tính đạo hàm của các hàm số sau:
a, \(y=\dfrac{2x-1}{x-1}\)
b, \(y=\dfrac{2x+1}{1-3x}\)
c, \(y=\dfrac{x^2+2x+2}{x+1}\)
d, \(y=\dfrac{2x^2}{x^2-2x-3}\)
e, \(y=x+1-\dfrac{2}{x-1}\)
g, \(y=\dfrac{2x^2-4x+5}{2x+1}\)
2. Tính đạo hàm của các hàm số sau:
a, \(y=\left(x^2+x+1\right)^4\)
b, y= (1-2x2)5
c, \(y=\left(\dfrac{2x+1}{x-1}\right)^3\)
d, \(y=\dfrac{\left(x+1\right)^2}{\left(x-1\right)^3}\)
e, \(y=\dfrac{1}{\left(x^2-2x+5\right)^2}\)
f, \(y=\left(3-2x^2\right)^4\)
a. \(y'=\dfrac{-1}{\left(x-1\right)}\)
b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)
c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)
d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)
e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)
g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
2.
a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)
b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)
c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)
d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)
e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)
f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
1. TÌm GTNN:
a, M=\(\dfrac{x^4+1}{\left(x^2+1\right)^2}\)
b, N=\(\dfrac{x^2}{-4y^2+20xy-29x^2}\)
2. Tìm GTNN và GTLN của biểu thức:
a,A=\(\dfrac{2x^2-2x+9}{x^2+2x+5}\)
b, B=\(\dfrac{4x^3}{x^2+1}\)
c, C=\(\dfrac{2\left(x^2+x+1\right)}{x^2+1}\)
d, D=\(\dfrac{x^2+xy+y^2}{x^2+y^2}\)với x khác 0
Thực hiện phép tính:
a) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
b) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
c) \(\dfrac{xy}{ab}+\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}-\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)
d) \(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)