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Câu a :
Ta có : \(\sqrt{5+3x}-\sqrt{5-3x}=a\)
\(\Leftrightarrow\left(\sqrt{5+3x}-\sqrt{5-3x}\right)^2=a^2\)
\(\Leftrightarrow5+3x-2\sqrt{\left(5+3x\right)\left(5-3x\right)}+5-3x=a^2\)
\(\Leftrightarrow10-2\sqrt{25-9x^2}=a^2\)
\(\Leftrightarrow2\sqrt{25-9x^2}=10-a^2\)
\(\Leftrightarrow\sqrt{25-9x^2}=\dfrac{10-a^2}{2}\)
\(\Leftrightarrow25-9x^2=\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow9x^2=25-\dfrac{\left(a^2-10\right)^2}{2}\)
\(\Leftrightarrow3x=\sqrt{\dfrac{50-\left(a^2-10\right)^2}{2}}\)
\(\Leftrightarrow x=\dfrac{\sqrt{50-\left(a^2-10\right)^2}}{3\sqrt{2}}\)
\(P=\dfrac{3\sqrt{2}.\sqrt{10+2\sqrt{\dfrac{10-a^2}{2}}}}{\sqrt{50-\left(a^2-10\right)^2}}\)
Bạn tự rút gọn nữa nhé :))
Câu b : \(M=\dfrac{2x+y+z-15}{x}+\dfrac{x+2y+z-15}{y}+\dfrac{x+y+2z-24}{z}\)
\(=\dfrac{x-3}{x}+\dfrac{y-3}{y}+\dfrac{z-12}{z}\)
\(=3-3\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{4}{z}\right)\le3-3\left[\dfrac{\left(1+1+2\right)^2}{12}\right]=-1\)
2
\(A=\sqrt{1-6x+9x^2}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{9x^2-6x+1}+\sqrt{9x^2-12x+4}\)
A= \(\sqrt{\left(3x-1\right)^2}+\sqrt{\left(3x-2\right)^2}=\left|3x-1\right|+\left|3x-2\right|\)
ta có |3x-1|+|3x-2|=|3x-1|+|2-3x| ≥ |3x-1+2-3x|=1
=> A ≥ 1
=> Min A =1 khi 1/3 ≤ x ≤ 2/3
Bài 1:
a: \(A=\left|2a-1\right|-2a\)
TH1: a>=1/2
A=2a-1-2a=-1
TH2: a<1/2
A=1-2a-2a=1-4a
b: \(B=x-2y-\left|x-2y\right|\)
TH1: x>=2y
A=x-2y-x+2y=0
TH2: x<2y
A=x-2y+x-2y=2x-4y
c: \(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
\(A=x^2+x^2-4=2x^2-4\)
TH2: -2<x<2
\(A=x^2+4-x^2=4\)
d: \(D=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
TH1: x>5
\(D=2x-1-1=2x-2\)
TH2: x<5
D=2x-1+1=2x
ta có:\(P=\sum\dfrac{y^2z^2}{x\left(y^2+z^2\right)}=\sum\dfrac{\dfrac{1}{x}}{\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)
đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\)thì giả thiết trở thành : \(a^2+b^2+c^2=1\).tìm Min \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{a^2+c^2}+\dfrac{c}{a^2+b^2}\)
ta có:\(\dfrac{a}{b^2+c^2}=\dfrac{a}{1-a^2}=\dfrac{a^2}{a\left(1-a^2\right)}\)
Áp dụng bất đẳng thức cauchy:
\(\left[a\left(1-a^2\right)\right]^2=\dfrac{1}{2}.2a^2\left(1-a^2\right)\left(1-a^2\right)\le\dfrac{1}{54}\left(2a^2+1-a^2+1-a^2\right)^3=\dfrac{4}{27}\)
\(\Rightarrow a\left(1-a^2\right)\le\dfrac{2}{3\sqrt{3}}\)\(\Rightarrow\dfrac{a^2}{a\left(1-a^2\right)}\ge\dfrac{3\sqrt{3}}{2}a^2\)
tương tự với các phân thức còn lại ta có:
\(P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\)
hay \(x=y=z=\sqrt{3}\)
Đặt \(\left\{{}\begin{matrix}\dfrac{1}{x}=a\\\dfrac{1}{y}=b\\\dfrac{1}{z}=c\end{matrix}\right.\) Thì bài toán trở thành
Cho \(a^2+b^2+c^2=1\) tính GTNN của \(P=\dfrac{a}{b^2+c^2}+\dfrac{b}{c^2+a^2}+\dfrac{c}{a^2+b^2}\)
Ta có:
\(a^2+b^2+c^2=1\)
\(\Rightarrow a^2+b^2=1-c^2\)
\(\Rightarrow\dfrac{c}{a^2+b^2}=\dfrac{c^2}{c\left(1-c^2\right)}\)
Mà ta có: \(2c^2\left(1-c^2\right)\left(1-c^2\right)\le\dfrac{\left(2c^2+1-c^2+1-c^2\right)^3}{27}=\dfrac{8}{27}\)
\(\Rightarrow c\left(1-c^2\right)\le\dfrac{2}{3\sqrt{3}}\)
\(\Rightarrow\dfrac{c^2}{c\left(1-c^2\right)}\ge\dfrac{3\sqrt{3}c^2}{2}\)
\(\Rightarrow\dfrac{c}{a^2+b^2}\ge\dfrac{3\sqrt{3}c^2}{2}\left(1\right)\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{b}{c^2+a^2}\ge\dfrac{3\sqrt{3}b^2}{2}\left(2\right)\\\dfrac{a}{b^2+c^2}\ge\dfrac{3\sqrt{3}a^2}{2}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) \(\Rightarrow P\ge\dfrac{3\sqrt{3}}{2}\left(a^2+b^2+c^2\right)=\dfrac{3\sqrt{3}}{2}\)
Dấu = xảy ra khi \(a=b=c=\dfrac{1}{\sqrt{3}}\) hay \(x=y=z=\sqrt{3}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\\ \Leftrightarrow\dfrac{x+y}{xy}+\left(\dfrac{1}{z}-\dfrac{1}{x+y+z}\right)=0\\ \Leftrightarrow\dfrac{x+y}{xy}+\dfrac{x+y}{z\left(x+y+z\right)}=0\\ \Leftrightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}\right)=0\\ \)
Nếu x+y=0 => x=-y
Nếu
\(\dfrac{1}{xy}+\dfrac{1}{xz+yz+z^2}=0\\ \Rightarrow xz+yz+z^2+xy=0\\ \Rightarrow\left(x+z\right)\left(y+z\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-z\\y=-z\end{matrix}\right.\)
Tự thế vào :v
Từ \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{x+y+z}\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}-\dfrac{1}{x+y+z}=0\)
\(\Rightarrow\dfrac{x+y}{xy}+\dfrac{x+y+z-z}{z\left(x+y+z\right)}=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{1}{xy}+\dfrac{1}{z\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(\dfrac{zx+zy+z^2+xy}{xyz\left(x+y+z\right)}\right)=0\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)
Ta có: x8 - y8 = (x + y)(x - y)(x2 + y2)(x4 + y4)
y9 + z9 = (y + z)(y8 - y7z + y6z2 - ... + z8)
z10 - x10 = (z + x)(z4 - z3x + z2x2 - zx3 + z4)(z5 - x5)
Vậy M = \(\dfrac{3}{4}\) + (x + y)(y + z)(z + x) = \(\dfrac{3}{4}\)
Bài này t đăng làm đc rồi, zô mà xem
Sau khi m ns thì t cx lm xog r :)