Nguyễn Văn Đạt

$ĐKXĐ : $ $a,b$ \(\ge0\)

Khi $b=0$ và $P=2$ thì :

$2 = \dfrac{\sqrt[]{2a}{a+2}$

$\to \sqrt[]{2a} = 2.(a+2)$

$\to 2a = 4.(a+2)^2$

$\to 4a^2+14a+16=0$

$\to 2a^2+7a+8=0$

$\denlta = 7^2-4.2.8 < 0$ nên pt vô nghiệm

Do đó không có $a$ thỏa mãn.

\(4x^2-\left(2+\sqrt{3}\right)^2=0\)

\(\Leftrightarrow\left(2x\right)^2=\left(2+\sqrt{3}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=2+\sqrt{3}\\2x=-2-\sqrt{3}\end{matrix}\right.\)

\(\Leftrightarrow x=\pm\frac{2+\sqrt{3}}{2}\)

\(\sqrt{6+\sqrt{11}}+\sqrt{6-\sqrt{11}}+3\sqrt{2}\)

\(=\frac{\sqrt{12+2\sqrt{11}}}{\sqrt{2}}+\frac{\sqrt{12-2\sqrt{11}}}{\sqrt{2}}+3\sqrt{2}\)

\(=\frac{\sqrt{\left(\sqrt{11}+1\right)^2}}{\sqrt{2}}+\frac{\sqrt{\left(\sqrt{11}-1\right)^2}}{\sqrt{2}}+3\sqrt{2}\)

\(=\frac{\sqrt{11}+1}{\sqrt{2}}+\frac{\sqrt{11}-1}{\sqrt{2}}+3\sqrt{2}\)

\(=\frac{2\sqrt{11}}{\sqrt{2}}+3\sqrt{2}=\sqrt{22}+3\sqrt{2}\)

a) $ĐKXĐ : x \neq 0$

Ta có :

\(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)

\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)

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

\(=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|\)

Với \(x< 0\) thì : \(P=\frac{x^2+3}{-x}+2-x=\frac{2x^2+3-2x}{-x}\)

Với \(0< x< 2\) thì : \(P=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)

Với \(x\ge2\) thì : \(P=\frac{x^2+3}{x}+x-2=\frac{2x^2+3-2x}{x}\)

Để $P$ nguyên thì : $x^2+3 \vdots x$

$\to x \in Ư(3)$

$ĐKXĐ : x \neq 2, x \neq 0, x \neq -2$

Ta có : $\dfrac{4x-2x^2}{x^3-4x} = \dfrac{1}{2}$

$\to \dfrac{2x.(x-2)}{x.(x-2).(x+2)} = \dfrac{1}{2}$

$\to \dfrac{2}{x+2} = \dfrac{1}{2} = \dfrac{2}{4}$

$\to x+2=4$

$\to x=2$ ( Không thỏa mãn $ĐKXĐ$ )

Vậy pt vô nghiệm.

$ĐKXĐ : $ $ a \neq 1, $ \(a\ge0\)

Ta có :

\(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)

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

\(=\frac{1+\sqrt{a}-a\sqrt{a}-a}{1-\sqrt{a}}\cdot\frac{\left(1-\sqrt{a}\right)^2}{\left[\left(1-\sqrt{a}\right).\left(1+\sqrt{a}\right)\right]^2}\)

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

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

\(=1\)

\(16x^2-9y^2+8x+1\)

\(=\left(4x+1\right)^2-9y^2\)

\(=\left(4x+3y+1\right)\left(4x+1-3y\right)\)

Xuất dữ liệu $a*b$ thì bạn không cần ghi kiểu dữ liệu.

Bạn chỉ cần nhập qua câu lệnh sau :

writeln(a*b);

Hoặc nếu bạn biết a,b là kiểu số gì rồi thì bạn có thể gán một biến t:=a*b;

Lưu ý t nguyên khi a,b cùng nguyên, nên khai báo là nguyên; còn nếu a,b là real thì t cũng là real.

Ta có : \(a=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}\)

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

Tương tự ta có \(b=\sqrt{3}-1\)

Thiết lập được : \(\sqrt{ab}=\sqrt{\left(\sqrt{3}+1\right).\left(\sqrt{3}-1\right)}=\sqrt{3-1}=\sqrt{2}\)

\(a+b=\sqrt{3}+1+\sqrt{3}-1=2\sqrt{3}\)

Khi đó : \(A=\frac{\sqrt{3}+1}{\sqrt{2}+\sqrt{3}-1}+\frac{\sqrt{3}-1}{\sqrt{2}-\sqrt{3}-1}-\frac{2\sqrt{3}}{\sqrt{2}}\)

......

Xét \(A=1+2+2^2+2^3+....+2^{2008}\)

\(\Leftrightarrow2A=2+2^2+2^3+...+2^{2009}\)

\(\Leftrightarrow2A-A=2^{2009}-1\)

Nên \(B=1\)

Xét \(B^2=6-x+x+2+2\sqrt{\left(6-x\right)\left(x+2\right)}\)

\(=8+2\sqrt{\left(6-x\right)\left(x+2\right)}\le8+\left(6-x+x+2\right)=16\)

\(\Leftrightarrow-4\le B\le4\)

Dấu "=" xảy ra \(\Leftrightarrow6-x=x+2\Leftrightarrow x=2\)

Vậy \(B_{max}=4\) khi $x=2$

ĐKXĐ : \(x\le1\)

Pt đã cho tương đương :

\(\sqrt{4\left(1-x\right)^2}=6\)

\(\Leftrightarrow4.\left(1-x\right)^2=36\)

\(\Leftrightarrow\left(x-1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\) Mà \(x\le1\)

$\to x=-2$ thỏa mãn đề.

a) \(x^2+y^2-z^2+2xy-2z-1\)

\(=\left(x^2+y^2+2xy\right)-\left(z^2+2z+1\right)\)

\(=\left(x+y\right)^2-\left(z+1\right)^2\)

\(=\left(x+y+z+1\right)\left(x+y-z-1\right)\)

b) \(\left(a+b+c\right)^2+\left(a-b+c\right)^2-4b^2\)

\(=\left(a+b+c\right)^2+\left(a-b+c-2b\right).\left(a-b+c+2b\right)\)

\(=\left(a+b+c\right)^2+\left(a-3b+c\right)\left(a+b+c\right)\)

\(=\left(a+b+c\right)\left(a+b+c+a-3b+c\right)\)

\(=\left(a+b+c\right)\left(2a-2b+2c\right)\)

\(=2.\left(a+b+c\right)\left(a-b+c\right)\)

Ta có : \(5x^2+2y^2-6xy+16x-8y+16=0\)

\(\Leftrightarrow10x^2+4y^2-12xy+32x-16y+32=0\)

\(\Leftrightarrow\left(4y^2-2.2y.3x+9x^2\right)+x^2+32x-16y+32=0\)

\(\Leftrightarrow\left(2y-3x\right)^2-2.4.\left(2y-3x\right)+16+x^2+8x+16=0\)

\(\Leftrightarrow\left(2y-3x-4\right)^2+\left(x+4\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(2y-3x-4\right)^2=0\\\left(x+4\right)^2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-4\\y=-4\end{matrix}\right.\)

Vậy \(\left(x,y\right)=\left(-4,4\right)\)

a) \(\left(3x^2-2xy+y^2\right)+\left(x^2-xy+2y^2\right)-\left(4x^2-y^2\right)\)

\(=3x^2-2xy+y^2+x^2-xy+2y^2-4x^2+y^2\)

\(=4y^2-3xy\)

b) \(\left(x^2-y^2+2xy\right)-\left(x^2+xy+2y^2\right)+\left(4xy-1\right)\)

\(=x^2-y^2+2xy-x^2-xy-2y^2+4xy-1\)

\(=-3y^2+5xy-1\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-5\ge0\\5-x\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\le5\end{matrix}\right.\) \(\Leftrightarrow x=5\)

Khi \(x=5\) thì pt đã cho vô nghiệm

Vậy pt vô nghiệm.

P/s : Không chắc chắn ...

a) \(\frac{1}{5}x^2y-10x^2y-\frac{1}{5}x^2y\)

\(=-10x^2y=-10.0,5^2.2=-5\)

b) \(5x^2y-7xy^2+5x^2y-10x^2y+5xy^2\)

\(=-2xy^2=-2.0,5.4=-44\)

ĐK : \(\left|x\right|\ge3\)

\(\sqrt{x^2-9}+\sqrt{x^2-6x+9}=0\)

\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}+\sqrt{\left(x-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{x-3}.\left(\sqrt{x+3}+\sqrt{x-3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}+\sqrt{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\) ( Thỏa mãn )

a) Xét \(\Delta=4-4.\left(-m\right)\)

Vì pt có nghiệm kép nên \(\Delta=0\Leftrightarrow m=-1\)

b) Xét \(\Delta=\left(-3\right)^2-4m=9-4m\)

Vì pt có nghiệm kép nên \(\Delta=0\Leftrightarrow m=\frac{9}{4}\)

uses crt;
var n,m,i : int64;
begin clrscr;
write('Nhap n:'); readln(n);
m:=n;i:=0;
while m<>0 do begin
m:=m div 10;
i:=i+1;
end;
write('So chu so cua ',n,' la :', i);
readln
end.

a) $ĐKXĐ : x \neq 1, x ≥ 0$

Ta có :

$M= \bigg(\dfrac{x+2}{x\sqrt[]{x}-1}+\dfrac{\sqrt[]{x}}{x+\sqrt[]{x}+1} - \dfrac{1}{\sqrt[]{x}-1}\bigg) . \dfrac{2}{\sqrt[]{x}-1}$

$ = \bigg(\dfrac{x+2+\sqrt[]{x}.(\sqrt[]{x}-1) - (x+\sqrt[]{x}+1)}{(\sqrt[]{x}-1).(x+\sqrt[]{x}+1)}\bigg).\dfrac{2}{\sqrt[]{x}-1}$

$ = \dfrac{x-2\sqrt[]{x}+1}{(\sqrt[]{x}-1).(x+\sqrt[]{x}+1)} . \dfrac{2}{\sqrt[]{x}-1}$

$ = \dfrac{(\sqrt[]{x}-1)^2}{(\sqrt[]{x}-1).(x+\sqrt[]{x}+1)} . \dfrac{2}{\sqrt[]{x}-1}$

$ = \dfrac{2}{x+\sqrt[]{x}+1}$

Vậy $M = \dfrac{2}{x+\sqrt[]{x}+1}$ với $x \neq 1, x ≥ 0 $

b) Ta có : $x=6+2\sqrt[]{5}$

$ = (\sqrt[]{5})^2+2.\sqrt[]{5}.1+1$

$ = (\sqrt[]{5}+1)^2$

$\to \sqrt[]{x} = \sqrt[]{5}+1$

Do đó : $M = \dfrac{2}{(\sqrt[]{5}+1)^2+\sqrt[]{5}+1+1}$

$ = \dfrac{2}{5+3\sqrt[]{5}+1}$

c) Ta thấy : $x+\sqrt[]{x} + 1 ≥ 1> 0 $

$\to \dfrac{2}{x+\sqrt[]{x}+1} ≤ 2$

Dấu "=" xảy ra $⇔x=0$

Vậy $M_{max} = 2$ tại $x=0$

ĐKXĐ : \(x\ge-1,y\ge-2,z\ge-3\)

Ta có :

\(x+y+z=2.\left(2\sqrt{x+1}+3\sqrt{y+2}+4\sqrt{z+3}\right)\)

\(\Leftrightarrow x-4\sqrt{x+1}+y-6\sqrt{y+2}+z-8\sqrt{z+3}=0\)

\(\Leftrightarrow\left(x+1\right)-2.\sqrt{x+1}.2+4+\left(y+2\right)-2.\sqrt{y+2}.3+9+\left(z+3\right)-2.\sqrt{z+3}.4+16=32\)

\(\Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(\sqrt{y+2}-3\right)^2+\left(\sqrt{z+3}-4\right)^2=32\)

Có cảm giác đề hơi sai hoặc mình sai :v