\(\sqrt{x^2-x}=\sqrt{3-x}\) 

ĐK: \(\left\{{}\begin{matrix}x^2-x\ge0\\3-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\ge1\\x\le0\end{matrix}\right.\\x\le3\end{matrix}\right.\) \(\Leftrightarrow x\le3\)  

\(\Leftrightarrow x^2-x=3-x\)

\(\Leftrightarrow x^2-x+x=3\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\left(tm\right)\)

Vậy: ... 

đkxđ: \(z\ge1;x\ge2;y\ge3\)

Đặt \(a=\sqrt{z-1}\ge0;b=\sqrt{x-2}\ge0;c=\sqrt{y-3}\ge0\)

\(\Rightarrow z=a^2+1;x=b^2+2;y=c^2+3\)

\(\Rightarrow A=\dfrac{a}{a^2+1}+\dfrac{b}{b^2+2}+\dfrac{c}{c^2+3}\)

Do các biến \(a,b,c\) độc lập nhau nên ta xét từng phân thức một.

Đặt \(f\left(a\right)=\dfrac{a}{a^2+1}\) \(\Rightarrow f\left(a\right).a^2-a+f\left(a\right)=0\) (*)

Nếu \(f\left(a\right)=0\) thì \(a=0\), rõ ràng đây không phải là GTLN cần tìm.

Xét \(f\left(a\right)\ne0\)

Để pt (*) có nghiệm thì \(\Delta=\left(-1\right)^2-4\left[f\left(a\right)\right]^2\ge0\) 

\(\Leftrightarrow\left(1+2f\left(a\right)\right)\left(1-2f\left(a\right)\right)\ge0\)

\(\Leftrightarrow-\dfrac{1}{2}\le f\left(a\right)\le\dfrac{1}{2}\)

\(f\left(a\right)=\dfrac{1}{2}\Leftrightarrow\dfrac{a}{a^2+1}=\dfrac{1}{2}\Leftrightarrow a^2+1=2a\Leftrightarrow a=1\) (nhận)

Vậy \(max_{f\left(a\right)}=\dfrac{1}{2}\).

 Tiếp đến, gọi \(g\left(b\right)=\dfrac{b}{b^2+2}\) \(\Rightarrow g\left(b\right).b^2-b+2g\left(b\right)=0\) (**)

 Tương tự nếu \(b=0\) thì vô lí. Xét \(b\ne0\). Khi đó để (**) có nghiệm thì \(\Delta=\left(-1\right)^2-8\left[g\left(b\right)\right]^2\ge0\)

\(\Leftrightarrow\left(1-2\sqrt{2}g\left(b\right)\right)\left(1+2\sqrt{2}g\left(b\right)\right)\ge0\)

\(\Leftrightarrow-\dfrac{1}{2\sqrt{2}}\le g\left(b\right)\le\dfrac{1}{2\sqrt{2}}\)

\(g\left(b\right)=\dfrac{1}{2\sqrt{2}}\Leftrightarrow\dfrac{b}{b^2+2}=\dfrac{1}{2\sqrt{2}}\Leftrightarrow b^2+2=2\sqrt{2}b\Leftrightarrow b=\sqrt{2}\) (nhận)

Vậy \(max_{g\left(b\right)}=\dfrac{1}{2\sqrt{2}}\)

Làm tương tự với \(h\left(c\right)=\dfrac{c}{c^2+3}\), ta được \(max_{h\left(c\right)}=\dfrac{1}{2\sqrt{3}}\), xảy ra khi \(c=\sqrt{3}\)

Vậy GTLN của A là \(\dfrac{1}{2}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{2\sqrt{3}}=\dfrac{6+3\sqrt{2}+2\sqrt{3}}{12}\), xảy ra khi \(\left(a,b,c\right)=\left(1,\sqrt{2},\sqrt{3}\right)\) hay \(\left(x,y,z\right)=\left(2,4,6\right)\).

Cái chỗ cuối mình sửa thế này nhé

C

c

XX được gọi là cặp tương đồng

XY được gọi là cặp không tương đồng 

vẽ trên phần mềm Paint nha bạn

Bài làm của bạn chưa đúng nhé:

Đến đoạn

$|x-3|=3-x$

$\Leftrightarrow |3-x|=3-x$
$\Leftrightarrow 3-x\geq 0$

$\Leftrightarrow x\leq 3$

Vậy pt có nghiệm $x\leq 3, x\in\mathbb{R}$

----------------------------

Bạn nhớ 1 tính chất này: Nếu $|a|=a$ thì $a\geq 0$.

\(\dfrac{x+2\sqrt{x}}{\sqrt{x}-1}=8\left(x\ge0;x\ne1\right)\)

\(\Leftrightarrow x+2\sqrt{x}=8\left(\sqrt{x}-1\right)\)

\(\Leftrightarrow x+2\sqrt{x}=8\sqrt{x}-8\)

\(\Leftrightarrow x+2\sqrt{x}-8\sqrt{x}+8=0\)

\(\Leftrightarrow x-6\sqrt{x}+8=0\)

\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\left(tm\right)\)

Vậy: ... 

\(\dfrac{x+2\sqrt{x}}{\sqrt{x}-1}=8\left(x\ge0,x\ne1\right)\\ < =>x+2\sqrt{x}=8\sqrt{x}-8\\ < =>x-6\sqrt{x}+8=0\\ < =>\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)=0\\ =>\left[{}\begin{matrix}\sqrt{x}-2=0\\\sqrt{x}-4=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=4\\x=16\end{matrix}\right.\left(TMDK\right)\)

\(=>S=\left\{4;16\right\}\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

\(=\frac{2}{\sqrt{x}-1}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+3-x}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

\(=\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

\(=\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)

=>\(2x-\sqrt{x}=\sqrt{x}+1\)

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}<1-\sqrt{x}\)

=>\(2\sqrt{x}-1<\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)=-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=-\left(x-1\right)=-x+1\)

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

\(=\frac{2}{\sqrt{x}-1}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2\left(\sqrt{x}+2\right)-5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+3-x}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

\(=\frac{x+\sqrt{x}-2+3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

\(=\frac{2\sqrt{x}-1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{\sqrt{x}}\)

=>\(2x-\sqrt{x}=\sqrt{x}+1\)

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

=>\(\frac{2\sqrt{x}-1}{\sqrt{x}+1}<1-\sqrt{x}\)

=>\(2\sqrt{x}-1<\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)=-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=-\left(x-1\right)=-x+1\)

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

CTM: \(R_1ntR_2\)

\(R_{tđ}=R_1+R_2\), mà \(R_2=3R_1\) nên:

\(R_{tđ}=R_1+3R_1=4R_1=17\) \(\Rightarrow R_1=\dfrac{17}{4}\Omega=4,25\Omega\)

\(\Rightarrow R_2=3R_1=3\cdot4,25=12,75\Omega\)

ctm là j bạn