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NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 3 2023
Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow xyz=1\) và \(x;y;z>0\)
Gọi biểu thức cần tìm GTNN là P, ta có:
\(P=\dfrac{1}{\dfrac{1}{x^3}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)}+\dfrac{1}{\dfrac{1}{y^3}\left(\dfrac{1}{z}+\dfrac{1}{x}\right)}+\dfrac{1}{\dfrac{1}{z^3}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)}\)
\(=\dfrac{x^3yz}{y+z}+\dfrac{y^3zx}{z+x}+\dfrac{z^3xy}{x+y}=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
\(P\ge\dfrac{\left(x+y+z\right)^2}{y+z+z+x+x+y}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}\)
\(P_{min}=\dfrac{3}{2}\) khi \(x=y=z=1\) hay \(a=b=c=1\)
22 tháng 3
Đặt \(a = \frac{1}{x} ; b = \frac{1}{y} ; c = \frac{1}{z} \Rightarrow x y z = 1\) và \(x ; y ; z > 0\)
Gọi biểu thức cần tìm GTNN là P, ta có:
\(P = \frac{1}{\frac{1}{x^{3}} \left(\right. \frac{1}{y} + \frac{1}{z} \left.\right)} + \frac{1}{\frac{1}{y^{3}} \left(\right. \frac{1}{z} + \frac{1}{x} \left.\right)} + \frac{1}{\frac{1}{z^{3}} \left(\right. \frac{1}{x} + \frac{1}{y} \left.\right)}\)
\(= \frac{x^{3} y z}{y + z} + \frac{y^{3} z x}{z + x} + \frac{z^{3} x y}{x + y} = \frac{x^{2}}{y + z} + \frac{y^{2}}{z + x} + \frac{z^{2}}{x + y}\)
\(P \geq \frac{\left(\left(\right. x + y + z \left.\right)\right)^{2}}{y + z + z + x + x + y} = \frac{x + y + z}{2} \geq \frac{3 \sqrt[3]{x y z}}{2} = \frac{3}{2}\)
\(P_{m i n} = \frac{3}{2}\) khi \(x = y = z = 1\) hay \(a = b = c = 1\)
S
subjects
CTVHS
31 tháng 8
theo đề ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\cdot\left(xy+yz+zx\right)=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\left(1\right)\)
ta co: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
mà x + y + z = 0
\(\Rightarrow x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\left(2\right)\)
a. VT = \(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+x^2z^2\right)\)
ta có: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)+2xyz\cdot\left(x+y+z\right)\)
vì x+y+z=0 nên: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)\)
từ (1) ta có: \(\left(x^2+y^2+z^2\right)^2=\left\lbrack-2\left(xy+yz+zx\right)^{}\right\rbrack^2\) (*)
\(=4\cdot\left(xy+yz+zx\right)^2=4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
ta có: \(4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
mà: \(2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4\)
thay vào (*) ta được:
\(\left(x^2+y^2+z^2\right)^2=\left(x^4+y^4+z^4\right)+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
\(=x^4+y^4+z^4+x^4+y^4+z^4=2\cdot\left(x^4+y^4+z^4\right)=VP\)
⇒ đpcm
b. \(VT=5\cdot\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)
\(=5\cdot\left(3xyz\right)\left(x^2+y^2+z^2\right)\)
\(=15xyz\cdot\left(x^2+y^2+z^2\right)\) (3)
\(x+y+z=0\Rightarrow x+y=-z\)
\(x^5+y^5+z^5=x^5+y^5+\left\lbrack-\left(x+y\right)\right\rbrack^5=x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5y^4+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)
\(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left\lbrack x^3+y^3+2xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left\lbrack\left(x+y\right)^3-3xy\left(x+Y\right)+2xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left\lbrack\left(x+Y\right)^3-xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left(x+Y\right)\left\lbrack\left(x+y\right)^2-xy\right\rbrack\)
vì x+y=-z nên ta có:
\(x^5+y^5+z^5=-5xy\left(-z\right)\left\lbrack\left(-z\right)^2-xy\right\rbrack=5xyz\left(x^2-zy\right)\)
mặt khác \(x+y=-z\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)
\(x^2+y^2+z^2=x^2+y^2+\left(x+y\right)^2\)
\(=x^2+y^2+x^2+2xy+y^2=2\cdot\left(x^2+xy+y^2\right)\)
\(z^2-xy=\left(x+y\right)^2-xy=x^2+2xy+y^2-xy=x^2+xy+y^2\)
vậy \(x^5+y^5+z^5=5xyz\cdot\left(x^2+xy+y^2\right)=\frac52xyz\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\cdot\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
⇒ \(6\cdot\left(x^5+y^5+z^5\right)=15xyz\left(x^2+y^2+z^2\right)\) (4)
từ (3) và (4) ⇒ VT = VP
TC
2
19 tháng 12 2023
Ta có
\(BC\perp AB';B'C'\perp AB'\) => BC//B'C'
\(\Rightarrow\dfrac{AB}{AB'}=\dfrac{BC}{B'C'}\Rightarrow\dfrac{x}{x+h}=\dfrac{a}{a'}\)
\(\Rightarrow a'x=ax+ah\Rightarrow x\left(a'-a\right)=ah\Rightarrow x=\dfrac{ah}{a'-a}\left(dpcm\right)\)
30 tháng 9 2024
Xét tam giác ABCABC có BC⊥ AB′BC⊥ AB′ và B′C′⊥AB′B′C′⊥AB′ nên suy ra BCBC // B′C′B′C′.
Theo hệ quả định lí Thalès, ta có: ABAB′ =BCBC′AB′AB =BC′BC
Suy ra xx+h =aa′x+hx =a′a
a′.x=a(x+h)a′.x=a(x+h)
a′.x−ax=aha′.x−ax=ah
x(a′−a)=ahx(a′−
S
subjects
CTVHS
31 tháng 8
theo đề ta có: \(x+y+z=0\Rightarrow\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2\cdot\left(xy+yz+zx\right)=0\)
\(\Rightarrow x^2+y^2+z^2=-2\left(xy+yz+xz\right)\left(1\right)\)
ta co: \(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
mà x + y + z = 0
\(\Rightarrow x^3+y^3+z^3-3xyz=0\Rightarrow x^3+y^3+z^3=3xyz\left(2\right)\)
a. VT = \(\left(x^2+y^2+z^2\right)^2=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+x^2z^2\right)\)
ta có: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)+2xyz\cdot\left(x+y+z\right)\)
vì x+y+z=0 nên: \(\left(xy+yz+zx\right)^2=\left(x^2y^2+y^2z^2+x^2z^2\right)\)
từ (1) ta có: \(\left(x^2+y^2+z^2\right)^2=\left\lbrack-2\left(xy+yz+zx\right)^{}\right\rbrack^2\) (*)
\(=4\cdot\left(xy+yz+zx\right)^2=4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
ta có: \(4\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
mà: \(2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)=x^4+y^4+z^4\)
thay vào (*) ta được:
\(\left(x^2+y^2+z^2\right)^2=\left(x^4+y^4+z^4\right)+2\cdot\left(x^2y^2+y^2z^2+z^2x^2\right)\)
\(=x^4+y^4+z^4+x^4+y^4+z^4=2\cdot\left(x^4+y^4+z^4\right)=VP\)
⇒ đpcm
b. \(VT=5\cdot\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)\)
\(=5\cdot\left(3xyz\right)\left(x^2+y^2+z^2\right)\)
\(=15xyz\cdot\left(x^2+y^2+z^2\right)\) (3)
\(x+y+z=0\Rightarrow x+y=-z\)
\(x^5+y^5+z^5=x^5+y^5+\left\lbrack-\left(x+y\right)\right\rbrack^5=x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5y^4+10x^3y^2+10x^2y^3+5xy^4+y^5\right)\)
\(=-5x^4y-10x^3y^2-10x^2y^3-5xy^4\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left\lbrack x^3+y^3+2xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left\lbrack\left(x+y\right)^3-3xy\left(x+Y\right)+2xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left\lbrack\left(x+Y\right)^3-xy\left(x+y\right)\right\rbrack\)
\(=-5xy\left(x+Y\right)\left\lbrack\left(x+y\right)^2-xy\right\rbrack\)
vì x+y=-z nên ta có:
\(x^5+y^5+z^5=-5xy\left(-z\right)\left\lbrack\left(-z\right)^2-xy\right\rbrack=5xyz\left(x^2-zy\right)\)
mặt khác \(x+y=-z\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)
\(x^2+y^2+z^2=x^2+y^2+\left(x+y\right)^2\)
\(=x^2+y^2+x^2+2xy+y^2=2\cdot\left(x^2+xy+y^2\right)\)
\(z^2-xy=\left(x+y\right)^2-xy=x^2+2xy+y^2-xy=x^2+xy+y^2\)
vậy \(x^5+y^5+z^5=5xyz\cdot\left(x^2+xy+y^2\right)=\frac52xyz\left(x^2+y^2+z^2\right)\)
\(\Rightarrow2\cdot\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
⇒ \(6\cdot\left(x^5+y^5+z^5\right)=15xyz\left(x^2+y^2+z^2\right)\) (4)
từ (3) và (4) ⇒ VT = VP
QT
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1
HQ
Hà Quang Minh
Giáo viên
10 tháng 9 2023
\({x^2} = {4^2} + {2^2} = 20 \Rightarrow x = 2\sqrt 5 \)
\({y^2} = {5^2} - {4^2} = 9 \Leftrightarrow y = 3\)
\({z^2} = {\left( {\sqrt 5 } \right)^2} + {\left( {2\sqrt 5 } \right)^2} = 25 \Rightarrow z = 5\)
\({t^2} = {1^2} + {2^2} = 5 \Rightarrow t = \sqrt 5 \)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 8
\(x+y+z=0\rArr\left(x+y+z\right)^2=0\)
\(\Rightarrow x^2+y^2+z^2+2xy+2yz+2zx=0\)
\(\rArr x^2+y^2+z^2+2\left(xy+yz+xz\right)=0\)
\(\rArr x^2+y^2+z^2=0\) (do \(xy+yz+xz=0\) )
\(\rArr x=y=z=0\)
Do đó:
\(\left(x-1\right)^{2023}+y^{2024}+\left(z+1\right)^{2025}=\left(0-1\right)^{2023}+0^{2024}+\left(0+1\right)^{2025}=-1+0+1=0\)
P
2
28 tháng 8
a:
b: TH1: \(\hat{BAD}>90^0;\hat{ABD}>90^0\)
Ta có: ABCD là hình thang
=>\(\hat{ABC}+\hat{BCD}=180^0\)
=>\(\hat{BCD}<180^0-90^0=90^0\)
=>\(\hat{BCD}<\hat{BAD}\)
TH2: \(\hat{ADC}>90^0;\hat{DCB}>90^0\)
Ta có: ABCD là hình thang
DC//AB
=>\(\hat{CDA}+\hat{DAB}=180^0\)
=>\(\hat{DAB}<180^0-90^0=90^0\)
=>\(\hat{DAB}<\hat{DCB}\)
c: Xét tứ giác ABCD có
AB//CD
AB=CD
Do đó: ABCD là hình bình hành
23 tháng 8
Bài 2:
a: ĐKXĐ: x∉{2;-2}
b: \(A=\frac{3x}{x-2}-\frac{2}{x+2}+\frac{2x-4}{x^2-4}\)
\(=\frac{3x}{x-2}-\frac{2}{x+2}+\frac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{3x}{x-2}-\frac{2}{x+2}+\frac{2}{x+2}=\frac{3x}{x-2}\)
c: Thay x=-5 vào A, ta được:
\(A=\frac{3\cdot\left(-5\right)}{-5-2}=\frac{-15}{-7}=\frac{15}{7}\)
d: Để A nguyên thì 3x⋮x-2
=>3x-6+6⋮x-2
=>6⋮x-2
=>x-2∈{1;-1;2;-2;3;-3;6-6}
=>x∈{1;2;4;0;5;-1;8;-4}
Kết hợp ĐKXĐ, ta được: x∈{1;4;0;5;-1;8;-4}
Bài 1:
a: \(A=x^2+10x+25\)
\(=x^2+2\cdot x\cdot5+5^2=\left(x+5\right)^2\)
b: \(B=x^2-y^2+8x-8y\)
=(x-y)(x+y)+8(x-y)
=(x-y)(x+y+8)
c: \(C=x^2+4x-5\)
\(=x^2+5x-x-5\)
=x(x+5)-(x+5)
=(x+5)(x-1)
BM
1
17 tháng 8
Bài 2:
a: \(\left(-\frac13x^2y\right)\cdot2xy^3=\left(-\frac13\cdot2\right)\cdot x^2\cdot x\cdot y\cdot y^3=-\frac23x^3y^4\)
b: \(\left(-\frac34x^2y\right)\cdot\left(-xy\right)^3=\left(-\frac34\right)\cdot\left(-1\right)\cdot x^2\cdot x^3\cdot y\cdot y^3=\frac34x^5y^4\)
c: \(\frac35\cdot x^2y^5\cdot x^3y^2\cdot\frac{-2}{3}=\left(\frac35\cdot\frac{-2}{3}\right)\cdot x^2\cdot x^3\cdot y^5\cdot y^2=-\frac25x^5y^7\)
d: \(\left(\frac34x^2y^3\right)\cdot\left(2\frac25x^4\right)=\frac34x^2y^3\cdot\frac{12}{5}x^4=\frac34\cdot\frac{12}{5}\cdot x^2\cdot x^4\cdot y^3=\frac95x^6y^3\)
e: \(\left(\frac{12}{15}x^4y^5\right)\cdot\left(\frac59x^2y\right)=\frac45\cdot\frac59\cdot x^4\cdot x^2\cdot y^5\cdot y=\frac49x^6y^6\)
f: \(\left(-\frac17x^2y\right)\left(-\frac{14}{5}x^4y^5\right)=\frac17\cdot\frac{14}{5}\cdot x^2\cdot x^4\cdot y\cdot y^5=\frac25x^6y^6\)
Bài 1: Các đơn thức là \(x^2y;-13;\left(-2\right)^3xy^7\)
a: \(VP=a^3+b^3+c^3-3bac\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=VT\)
b: \(VT=\left(3a+2b-1\right)\left(a+5\right)-2b\left(a-2\right)\)
\(=3a^2+15a+2ab+10b-a-5-2ab+4b\)
\(=3a^2+14a+14b-5\)
\(VP=\left(3a+5\right)\left(a+3\right)+2\left(7b-10\right)\)
\(=3a^2+9a+5a+15+14b-20\)
\(=3a^2+14a+14b-5\)
=>VT=VP
c: \(VT=a\left(b-x\right)+x\left(a+b\right)\)
\(=ab-ax+ax+bx\)
\(=ab+bx=b\left(a+x\right)=VP\)
d: \(VT=a\left(b-c\right)-b\left(a+c\right)+c\left(a-b\right)\)
\(=ab-ac-ab-bc+ca-cb\)
\(=-2bc\)
=VP