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\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
\(7\left(x-2004\right)^2=23-y^2\)
\(\Rightarrow7\left(x-2004\right)^2+y^2=23\left(1\right)\)
Vì \(y^2\ge0\) nên \(\left(x-2004\right)^2\le\frac{23}{7}\) suy ra \(\left[\begin{matrix}\left(x-2004\right)^2=0\\\left(x-2004\right)^2=1\end{matrix}\right.\)
*)Xét \(\left(x-2004\right)^2=0\) thay vào \((1)\) ta có: \(y^2=23\) (loại)
*)Xét \((x-2004)^2=1\) thay vào \((1)\) ta có \(y^2=16\)
Từ đó ta tìm được \(\left[\begin{matrix}\left\{\begin{matrix}x=2005\\y=4\end{matrix}\right.\\\left\{\begin{matrix}x=2003\\y=4\end{matrix}\right.\end{matrix}\right.\)
\(\frac{64}{\left(-2\right)^x}=\left(-16\right)^2:4^3\)
<=> \(\frac{64}{\left(-2\right)^x}=4\)
<=> \(\frac{64}{\left(-2\right)^x}=\frac{64}{16}\)
<=> (-2)x = 16
<=> x = 4
a,
\(\frac{x}{5}=\frac{y}{7}=\frac{z}{3}\)
Mà : x2+y2+z2=585
=> \(\frac{x^2}{25}=\frac{y^2}{49}=\frac{z^2}{9}\)
\(\Rightarrow\frac{x^2+y^2+z^2}{25+49+9}=\frac{585}{93}=\frac{195}{31}\)
=> x=195/31.5
=> y=195/31.7
=> z=195/31.3
Xong :)
!)
=> x(x - 1)=0
=> \(\left[\begin{array}{nghiempt}x=1\\x-1=0\end{array}\right.\)
=>\(\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
Vậy đa thức có nghiệm là x=0 ; x=1
1) \(x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-1=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=1\end{array}\right.\)
b) \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-2=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=2\end{array}\right.\)
c)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\end{array}\right.\)
d)\(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\3x-4=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{4}{3}\end{array}\right.\)
a) \(2x^2-4x+7\)
\(=2\left(x^2-2x+\dfrac{7}{2}\right)\)
\(=2\left(x^2-x-x+\dfrac{7}{2}\right)\)
\(=2\left(x^2-x-x+1+\dfrac{5}{2}\right)\)
\(=2\left[\left(x-1\right)^2+\dfrac{5}{2}\right]\)
\(=2\left(x-1\right)^2+5\)
Vì \(2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2+\dfrac{5}{2}\ge\dfrac{5}{2}>0\)
\(\Rightarrow\) đt vô nghiệm.
Mấy câu kia cũng tách tương tự.
" Giữ nguyên hạng tử bậc hai chia đội hạng tử bậc nhất cân bằng hệ số để đạt được tỉ lệ thức"
Chúc bạn học tốt!!!
a/ theo bài ra, ta có:
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=x+y+z\)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+z}{y+z+1+z+x+1+x+y-2}=\frac{x+y+z}{2\left(x+y+z\right)}=x+y+z\)
- nếu x+y+z = 0 => x = y= z = 0
- nếu x+y+z khác 0 => x+y+z = \(\frac{1}{2}\)
=> y + z = \(\frac{1}{2}\) - x
=> z + x = \(\frac{1}{2}\) - y
=> x + y = \(\frac{1}{2}\) - z
=> \(\frac{x}{\frac{1}{2}-x+1}=\frac{y}{\frac{1}{2}-y+1}=\frac{z}{\frac{1}{2}-z-2}=\frac{1}{2}\)
=> 2x = \(\frac{1}{2}\) - x + 1 => x = \(\frac{1}{2}\)
=> 2y = \(\frac{1}{2}-y+1\) => y = \(\frac{1}{2}\)
=> 2z = \(\frac{1}{2}-z-2\) => z = \(\frac{-1}{2}\)
vậy x = 0 hoặc 1/2
y = 0 hoặc 1/2
z = 0 hoặc -1/2
mk lm câu b bái 1 nha
Ta có: \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{4}\Rightarrow\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}=\frac{\left(2x-2\right)+\left(3y-6\right)-\left(z-3\right)}{4+9-4}\\=\frac{2x+3y-z-2-6+3}{9}=\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=\frac{45}{9}=5\)
Suy ra
x - 1 = 5 . 2 = 10
x = 10 + 1
→ x = 11
y - 2 = 3 . 5 = 15
y = 15 + 2
→ y = 17
z - 3 = 4 . 5 = 20
z = 20 + 3
→ z = 23
a/ \(\left(4x^2y^3\right)\left(x^ny^7\right)=4x^5y^{10}\)
\(\Leftrightarrow4x^{2+n}y^{3+7}=4x^5y^{10}\)
\(\Rightarrow2+n=5\Rightarrow n=3\)
Vậy \(n=3\)
b/ \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Leftrightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left[{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}n=5\\m=11\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}m=11\\n=5\end{matrix}\right.\)
a) \(\left(4x^2\times y^3\right)\left(x^n\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4\times\left(x^2\times x^n\right)\times\left(y^3\times y^7\right)=4x^5y^{10}\)
\(\Rightarrow4x^{2+x}y^{10}=4x^5y^{10}\)
\(\Rightarrow x^{2+n}=x^5\)
\(\Rightarrow2+n=5\)
\(\Rightarrow n=5-2\)
\(\Rightarrow n=3\)
Vậy \(n=3\).
b) \(\left(-7x^4y^m\right)\left(-5x^ny^4\right)=35x^9y^{15}\)
\(\Rightarrow\left[\left(-7\right)\times\left(-5\right)\right]\times\left(x^4\times x^n\right)\times\left(y^m\times y^4\right)=35x^9y^{15}\)
\(\Rightarrow35x^{4+n}y^{m+4}=35x^9y^{15}\)
\(\Rightarrow\left\{{}\begin{matrix}x^{4+n}=x^9\\y^{m+4}=y^{15}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}4+n=9\\m+4=15\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=9-4\\m=15-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n=5\\m=9\end{matrix}\right.\)
Vậy \(m=9\) và \(n=5\).
2/\(A=75\left(4^{2004}+4^{2003}+...+1\right)+25\)
\(=25\left(3\left(4^{2004}+4^{2003}+...+1\right)+1\right)\)
\(=25\left(3\left(4^{2004}+4^{2003}+...4\right)+4\right)\)
\(=25.4\left(3\left(4^{2003}+4^{2002}+...+1\right)+1\right)\)
\(=100\left(3\left(4^{2003}+4^{2002}+...+1\right)+1\right)\)
Vậy \(A⋮100\)
Bài 1:
a) \(xy+3x-y=6\)
\(\Rightarrow x\left(y+3\right)-y=6\)
\(\Rightarrow x\left(y+3\right)-\left(y+3\right)=3\)
\(\Rightarrow\left(x-1\right)\left(y+3\right)=3\)
Ta có bảng sau:
Vậy cặp số \(\left(x;y\right)\) là \(\left(2;0\right);\left(4;-2\right);\left(0;-6\right);\left(-2;-4\right)\)
c) \(\left|x+3\right|+\left|x+1\right|=3x\)
Mà \(\left|x+3\right|+\left|x+1\right|\ge0\)
\(\Rightarrow3x\ge0\Rightarrow x\ge0\)
\(\Rightarrow x+3+x+1=3x\)
\(\Rightarrow x=4\)
Vậy x = 4