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bai1
\(3a\left(2+b\right)-a\left(1-b\right)-4ab=6a+3ab-a+ab-4ab=5a=\frac{5}{229}\)
bai3
\(M=4\left(X-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)=\)
\(4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x=-24\)
bai 4
\(\text{a(x-y)+b(y-x)}=\left(x-y\right)\left(a-b\right)\)
bai 5
ta co cong thuc tinh tong 1+2+3+4+5+...+150=\(\frac{\left(1+150\right)150}{2}=11325\)
a11325
bai 6
\(p=x\left(5x+15y\right)-5y\left(3x-2y\right)-5y^2+10\)
\(=5x^2+15xy-15xy+10y^2-5y^2+10=5x^2+5y^2+10=5\left(x^2+y^2\right)+10\)
ta nhan thay rang de P=10 thi (x2+y2)=0 suy ra x=y=0
P=0 thi (x2+y2)= -2 ma so chinh phuong bao gioi cung lon hon 0 nen truong hop nay vo nghiem de thoa man
Bài 2:
a) ĐK: $x\geq \pm \frac{1}{2}; x\neq 0$
\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}=\frac{(2x+1)^2-(2x-1)^2}{(2x-1)(2x+1)}.\frac{10x-5}{4x}\)
\(\frac{4x^2+4x+1-(4x^2-4x+1)}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}=\frac{8x}{(2x-1)(2x+1)}.\frac{5(2x-1)}{4x}\)
\(=\frac{10}{2x+1}\)
b) ĐK : $x\neq 0;-1$
\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{x}+x-2\right)=\left(\frac{1}{x(x+1)}-\frac{x(2-x)}{x(x+1)}\right):\frac{1+x^2-2x}{x}\)
\(=\frac{1-2x+x^2}{x(x+1)}.\frac{x}{1+x^2-2x}=\frac{x}{x(x+1)}=\frac{1}{x+1}\)
Bài 3:
a) ĐKXĐ: \(x\neq \pm 1\)
b)
\(A=\left(\frac{x+1}{2x-2}-\frac{3}{1-x^2}-\frac{x+3}{2x+2}\right).\frac{4x^2-4}{5}\)
\(=\left[\frac{(x+1)^2}{2(x-1)(x+1)}+\frac{6}{2(x-1)(x+1)}-\frac{(x+3)(x-1)}{2(x+1)(x-1)}\right].\frac{4(x^2-1)}{5}\)
\(=\frac{(x+1)^2+6-(x^2+2x-3)}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}\)
\(=\frac{10}{2(x-1)(x+1)}.\frac{4(x-1)(x+1)}{5}=4\)
c) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\ \Leftrightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\\ \Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\\ \Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\\ \Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\\ \Leftrightarrow\left(x+2005\right)=0\Leftrightarrow x=-2005\)
câu egf làm tương tự
Ta có: \(\left(x+y\right)^2=\left(x-y\right)^2+4xy\)
Thay số ta được:
\(\left(x+y\right)^2=4^2+4.5\)
\(\Rightarrow\left(x+y\right)^2=16+20=36\)
\(\Rightarrow x+y=\sqrt{36}=6\)
Vậy: \(x+y=6\)
tu x-y=4suy ra y=x-4
thay vao xy=5suy ra x(x-4)=5
suy ra x^2-4x+4=9
suy ra (x-2)^2=9
suy ra x-2=+-3
vi x<0 suy ra x=-3+2=-1
suy ra y=x-4=-1-4=-5
suy ra x+y=-1+-5=-6
Bài 1:
a) Đặt \(a=\dfrac{1}{229},b=\dfrac{1}{433}\), ta được
\(M=3a\left(2+b\right)-a\left(1-b\right)-4ab\)
\(M=6a+3ab-a+ab-4ab\)
\(M=5a\)
b) Ta có:
\(M=5a\)
\(M=\dfrac{5}{229}\)
Bài 2:
\(x=16\)
\(\Rightarrow x+1=17\left(1\right)\)
Thay (1) vào P, ta được:
\(P=x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+x+1+3\)
\(P=x^4-x^4-x^3+x^3+x^2-x^2-x+x+1+3\)
\(P=4\)
Bài 3:
\(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x^2\left(x-1\right)\)
\(=4x-24-2x^2-3x^3+5x^2-4x+3x^3-3x^2\)
\(=-24\)
Vậy biểu thức không phụ thuộc vào x
Bài 4:
\(a\left(x-y\right)+b\left(y-x\right)\)
\(=a\left(x-y\right)-b\left(x-y\right)\)
\(=\left(x-y\right)\left(a-b\right)\)
Bài 5:
a) \(a.a^2.a^3.a^4.a^5a^6...a^{150}\)
\(=a^{1+2+3+4+5+6+...+150}\)
Đặt \(A=1+2+3+...+150\)
\(A=\dfrac{150-1+1}{2}\left(1+150\right)\)
\(A=75.151\)
\(A=2265\)
Vậy 1 + 2 + 3 +...+ 150 = 2265 (1)
Thay (1) vào ta được
\(a^{1+2+3+4+5+6+...+150}=a^{2265}\)
b) \(x^{2-k}.x^{1-k}.x^{2k-3}\)
\(=x^{2-k+1-k+2k-3}\)
\(=x^0\)
\(=1\)
Bài 6:
a) \(P=x\left(5x+15y\right)-5y\left(3x-2y\right)-5\left(y^2-2\right)\)
\(P=5x^2+15xy-15xy+10y^2-5y^2+10\)
\(P=5x^2+5y^2+10\)
b) \(P=0\)
\(\Rightarrow5x^2+5y^2+10=0\)
\(\Rightarrow5\left(x^2+y^2+2\right)=0\)
\(\Rightarrow x^2+y^2+2=0\)
\(\Rightarrow x^2+y^2=-2\)
Vì \(x^2\ge0\)
\(y^2\ge0\)
\(\Rightarrow x^2+y^2\ge0\)
Mà \(x^2+y^2=-2\)
=> Không tồn tại cặp số x và y để P = 0
\(P=10\)
\(\Rightarrow5x^2+5y^2+10=10\)
\(\Rightarrow5x^2+5y^2=0\)
\(\Rightarrow5\left(x^2+y^2\right)=0\)
\(\Rightarrow x^2+y^2=0\)
Vì \(x^2\ge0\) với mọi x
\(y^2\ge0\) với mọi y
\(\Rightarrow x^2+y^2\ge0\)
Mà \(x^2+y^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\y^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)