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Bài 1:
a) \(3x^2-2x(5+1,5x)+10=3x^2-(10x+3x^2)+10\)
\(=10-10x=10(1-x)\)
b) \(7x(4y-x)+4y(y-7x)-2(2y^2-3,5x)\)
\(=28xy-7x^2+(4y^2-28xy)-(4y^2-7x)\)
\(=-7x^2+7x=7x(1-x)\)
c)
\(\left\{2x-3(x-1)-5[x-4(3-2x)+10]\right\}.(-2x)\)
\(\left\{2x-(3x-3)-5[x-(12-8x)+10]\right\}(-2x)\)
\(=\left\{3-x-5[9x-2]\right\}(-2x)\)
\(=\left\{3-x-45x+10\right\}(-2x)=(13-46x)(-2x)=2x(46x-13)\)
Bài 2:
a) \(3(2x-1)-5(x-3)+6(3x-4)=24\)
\(\Leftrightarrow (6x-3)-(5x-15)+(18x-24)=24\)
\(\Leftrightarrow 19x-12=24\Rightarrow 19x=36\Rightarrow x=\frac{36}{19}\)
b)
\(\Leftrightarrow 2x^2+3(x^2-1)-5x(x+1)=0\)
\(\Leftrightarrow 2x^2+3x^2-3-5x^2-5x=0\)
\(\Leftrightarrow -5x-3=0\Rightarrow x=-\frac{3}{5}\)
\(2x^2+3(x^2-1)=5x(x+1)\)
b, \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left[\left(x-y\right)+\left(z-x\right)\right]+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)^2\left(x-y\right)-\left(y-z\right)^2\left(x-y\right)-\left(y-z\right)^2\left(z-x\right)+\left(z-x\right)^2\left(z-x\right)\)
\(=\left(x-y\right)\left[\left(x-y\right)^2-\left(y-z\right)^2\right]-\left(z-x\right)\left[\left(y-z\right)^2-\left(z-x\right)^2\right]\)
\(=\left(x-y\right)\left(x-y-y+z\right)\left(x-y+y-z\right)-\left(z-x\right)\left(y-z-z+x\right)\left(y-z+z-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(z-x\right)\left(y-2z+x\right)\left(y-x\right)\)
\(=\left(x-y\right)\left(x-2y+z\right)\left(x-z\right)-\left(x-z\right)\left(y-2z+x\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(x-2y+z-y+2z-x\right)\)
\(=\left(x-y\right)\left(x-z\right)\left(3z-3y\right)\)
\(=3\left(x-y\right)\left(x-z\right)\left(z-y\right)\)
c, \(x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left[\left(y-x\right)-\left(z-x\right)\right]-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-x\right)-y^2z^2\left(y-x\right)+y^2z^2\left(z-x\right)-z^2x^2\left(z-x\right)\)
\(=\left(x^2y^2-y^2z^2\right)\left(y-x\right)+\left(y^2z^2-z^2x^2\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)+z^2\left(y-x\right)\left(x+y\right)\left(z-x\right)\)
\(=y^2\left(x-z\right)\left(x+z\right)\left(y-x\right)-z^2\left(y-x\right)\left(x+y\right)\left(x-z\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[y^2\left(x+z\right)-z^2\left(x+y\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y^2x+y^2z-z^2x-z^2y\right)\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y^2-z^2\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left[x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\right]\)
\(=\left(x-z\right)\left(y-x\right)\left(y-z\right)\left(xy+xz+yz\right)\)
d, \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xyz-3xy\left(x+y\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
Có:\(x+y+z=0\)
\(\Rightarrow\left(x+y+z\right) ^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\)
Có:
\(\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2\)
\(=y^2-2yz+z^2+z^2-2xz+z^2+x^2-2xy+y^{^2}\)
\(=2\left(x^2+y^2+z^2\right)-2\left(xy+yz+xz\right)\)
\(=2\left(x^2+y^2+z^2\right)+x^2+y^2+z^2\)
\(=3\left(x^2+y^2+z^2\right)\)
\(\Rightarrow\dfrac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}\)
\(=\dfrac{x^2+y^2+z^2}{3\left(x^2+y^2+z^2\right)}\)
\(=\dfrac{1}{3}\)
Lời giải:
Đặt \(B=\frac{x^2+y^2+z^2}{(y-z)^2+(z-x)^2+(x-y)^2}\)
\(\Leftrightarrow B=\frac{x^2+y^2+z^2}{2(x^2+y^2+z^2-xy-yz-xz)}\)
Lại có
\(x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)\)
\(\Rightarrow B=\frac{-2(xy+yz+xz)}{2[-2(xy+yz+xz)-(xy+yz+xz)]}=\frac{-2(xy+yz+xz)}{-6(xy+yz+xz)}=\frac{1}{3}\)
1, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1), (2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\) \(\xrightarrow[]{}\) đpcm
5. a, Ta có: \(\left(x-1\right)^2\ge0\Leftrightarrow x^2-2x+1\ge0\Leftrightarrow x^2+1\ge2x\) (1)
\(\left(y-1\right)^2\ge0\Leftrightarrow y^2-2y+1\ge0\Leftrightarrow y^2+1\ge2y\) (2)
\(\left(z-1\right)^2\ge0\Leftrightarrow z^2-2z+1\ge0\Leftrightarrow z^2+1\ge2z\) (3)
Từ (1),(2) và (3) suy ra:
\(x^2+1+y^2+1+z^2+1\ge2x+2y+2z\)
<=> \(x^2+y^2+z^2+3\ge2\left(x+y+z\right)\)
mà x+y+z=3
=>\(x^2+y^2+z^2+3\ge2.3=6\)
<=> \(x^2+y^2+z^2\ge6-3=3\)
<=> \(A\ge3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTNN của A=x2+y2+z2 là 3 khi x=y=z=1
b, Ta có: x+y+z=3
=> \(\left(x+y+z\right)^2=9\)
<=> \(x^2+y^2+z^2+2xy+2yz+2xz=9\)
<=> \(x^2+y^2+z^2=9-2xy-2yz-2xz\)
mà \(x^2+y^2+z^2\ge3\) (theo a)
=> \(9-2xy-2yz-2xz\ge3\)
<=> \(-2\left(xy+yz+xz\right)\ge3-9=-6\)
<=> \(xy+yz+xz\le\dfrac{-6}{-2}=3\)
<=> \(B\le3\)
Dấu "=" xảy ra khi x=y=z=1
Vậy GTLN của B=xy+yz+xz là 3 khi x=y=z=1
a) A=(x+z)(y+t)
= xy+xt+zy+zt
Áp dụng bất đẳng thức cô si cho 2 số ta có
x2+y2 ≥ 2\(\sqrt{x^2y^2}\)
⇔x2+y2 ≥ 2xy
TT ta có
x2+t2 ≥ 2xt
y2+z2 ≥ 2yz
z2+t2 ≥ 2zt
cộng vế vs vế ta có
=> x2+y2+x2+t2+y2+z2+t2 ≥ 2xy+2xt+2yz+2zt
⇔ 2(x2+y2+z2+t2) ≥ 2(xy+xt+yz+zt)
⇔ 2 .1 ≥2 A
⇔ 1≥ A
⇔ A ≤ 1
=> Max A =1 dấu "=" xảy ra khi x=y=t=z= \(\pm\dfrac{1}{2}\)
Câu b)
Đây là bài toán quen thuộc của dạng toán xác định điểm rơi trong BĐT Cô-si:
Áp dụng BĐT Cô-si:
\(\frac{2}{3}x^2+\frac{2}{3}y^2\geq 2\sqrt{\frac{2}{3}x^2.\frac{2}{3}y^2}=\frac{4}{3}|xy|\geq \frac{4}{3}xy\)
\(\frac{1}{3}x^2+\frac{4}{3}t^2\geq 2\sqrt{\frac{1}{3}x^2.\frac{4}{3}t^2}=\frac{4}{3}|xt|\geq \frac{4}{3}xt\)
\(\frac{1}{3}y^2+\frac{4}{3}z^2\geq 2\sqrt{\frac{1}{3}y^2.\frac{4}{3}z^2}=\frac{4}{3}|yz|\geq \frac{4}{3}yz\)
\(\frac{2}{3}z^2+\frac{2}{3}t^2\geq 2\sqrt{\frac{2}{3}z^2.\frac{2}{3}t^2}=\frac{4}{3}|zt|\geq \frac{4}{3}zt\)
Cộng theo vế các BĐT thu được và rút gọn:
\(\Rightarrow x^2+y^2+2z^2+2t^2\geq \frac{4}{3}(xy+xt+yz+zt)\)
\(\Leftrightarrow \frac{4}{3}(xy+xt+yz+zt)\leq 1\)
\(\Leftrightarrow B=(x+z)(y+t)\leq \frac{3}{4}\) hay $B_{\max}=\frac{3}{4}$
Dấu bằng xảy ra khi \(x=y=2z=2t\Leftrightarrow (x,y,z,t)=\left(\frac{1}{\pm \sqrt{3}}; \frac{1}{\pm\sqrt{3}}; \frac{1}{\pm 2\sqrt{3}}; \frac{1}{\pm 2\sqrt{3}}\right)\)