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4.Nếu\(|x-1|=0\)
thì x = 1.=> lx+2l = 3 và lx+3l = 4.
=>lx-1l+lx+2l+lx+3l=0+3+4=7.
Nếu \(|x+2|=0\)
thì x=-2 =>lx-1l=3 và lx+3l=1.
=>lx-1l+lx+2l+lx+3l=0+3+1=4.
Nếu \(|x+3|=0\)
thì x=-3 =>lx-1l=4 và lx+2l=1.
=>lx-1l+lx+2l+lx+3l=5.
Vậy \(Min_{\text{lx-1l+lx+2l+lx+3l}}=4\).
\(1.\) Với : x = 25 ( TM ĐKXĐ), thì : \(A=\dfrac{7}{\sqrt{25}+8}=\dfrac{7}{5+8}=\dfrac{7}{13}\)
2. \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-24}{x-9}=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)3. \(P=A.B=\dfrac{7}{\sqrt{x}+8}.\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{7}{\sqrt{x}+3}\)
Để P ∈ Z thì : \(\sqrt{x}+3\) ∈ Ư(7)
+) \(\sqrt{x}+3=7\) ⇔\(x=16\) ( TM ĐK)
+) \(\sqrt{x}+3=-7\) ⇔ Vô nghiệm
+) \(\sqrt{x}+3=1\)⇔ Vô nghiệm
+) \(\sqrt{x}+3=-1\) ⇔ Vô nghiệm
KL...............
1) \(4a\left(x-5\right)-2\left(5-x\right)\)
\(=4a\left(x-5\right)+2\left(x-5\right)\)
\(=2\left(x-5\right)\left(2a+1\right)\)
2) \(-3a\left(x-3\right)-a^2\left(3-x\right)\)
\(=-3a\left(x-3\right)+a^2\left(x-3\right)\)
\(=a\left(x-3\right)\left(-3+a\right)\)
3) \(2a^2b\left(x+y\right)-4a^3b\left(-x-y\right)\)
\(=2a^2b\left(x+y\right)+4a^3b\left(x+y\right)\)
\(=2a^2b\left(x+y\right)\left(1+2a\right)\)
4) \(-3a\left(x-3\right)-a^2\left(3-a\right)\)
Mình nghĩ câu này đề sai và hình như nó là câu 2 thì phải
5) \(x^{m+1}-x^m\)
\(=x^m.x-x^m\)
\(=x^m\left(x-1\right)\)
6) \(x^{m+1}+x^m\)
\(=x^m.x+x^m\)
\(=x^m\left(x+1\right)\)
7) \(x^{m+2}-x^m\)
\(=x^m.x^2-x^m\)
\(=x^m\left(x^2-1\right)\)
\(=x^m\left(x-1\right)\left(x+1\right)\)
8) \(x^{m+2}-x^2\)
\(=x^m.x^2-x^2\)
\(=x^2\left(x^m-1\right)\)
9) \(x^{m+2}-x^{m+1}\)
\(=x^{m+1}.x-x^{m+1}\)
\(=x^{m+1}\left(x-1\right)\)
\(\dfrac{4^x}{2^{x+y}}=8\Rightarrow\dfrac{2^{2x}}{2^{x+y}}=2^3\\ \Rightarrow2^{2x-x-y}=2^3\Rightarrow x-y=3\left(1\right)\)
\(\dfrac{9^{x+y}}{3^{5y}}=243\Rightarrow\dfrac{3^{2x+2y}}{3^{5y}}=3^5\\ \Rightarrow3^{2x+2y-5y}=3^5\Rightarrow2x-3y=5\left(2\right)\)
từ (1) và (2) ta có hpt:
\(\left\{{}\begin{matrix}x-y=3\\2x-3y=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\y=1\end{matrix}\right.\)
vậy cặp số nguyên x,y là 4 và 1
\(a,ĐK:x\ne-3;x\ne-2\)
\(b,ĐK:x\ne3\)
\(c,ĐK:x\ne0;x\ne\frac{4}{3}\)
\(d,ĐK:x\ne-2;e,ĐK:x\ne\pm2;f:ĐK:x\ne2\)
\(A=0\Leftrightarrow2x+6=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\left(loai\right)\)
\(B=0\Leftrightarrow x^2-9=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\left(loai\right)\\x=-3\end{matrix}\right.\Leftrightarrow x=-3\) \(C=0\Leftrightarrow9x^2-16=0\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{4}{3}\left(loai\right)\\x=-\frac{4}{3}\end{matrix}\right.\Leftrightarrow x=-\frac{4}{3}\) \(D=0\Leftrightarrow\left(x+2\right)=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\left(loai\right)\)
\(E=0\Leftrightarrow x\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\left(loai\right)\end{matrix}\right.\Leftrightarrow x=0\)
\(F=0\Leftrightarrow x^2+2x+4=0\Leftrightarrow\left(x+1\right)^2+3=0\left(voli\right)\)
1.
a)
\(5x\left(x-2y\right)+2\left(2y-x\right)^2\\ =5x\left(x-2y\right)+2\left(x-2y\right)^2\\ =\left(x-2y\right)\left[5x+2\left(x-2y\right)\right]\\ =\left(x-2y\right)\left(5x+2x-4y\right)\\ =\left(x-2y\right)\left(7x-4y\right)\)
b)
\(7x\left(y-4\right)^2-\left(4-y\right)^3\\ =7x\left(4-y\right)^2-\left(4-y\right)^3\\ =\left(4-y\right)^2\left(7x+y-4\right)\)
c)
\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\\ =\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\\ =\left(4x-8\right)\left(x^2+6-x-7-9\right)\\ =4\left(x-2\right)\left(x^2-x-10\right)\)
a) \(ĐKXĐ:x\ne\pm3\)
\(M=\frac{3+x}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\)
\(\Leftrightarrow M=\frac{\left(x+3\right)^2-18+\left(x-3\right)^2}{x^2-9}\)
\(\Leftrightarrow M=\frac{x^2+6x+9-18+x^2-6x+9}{x^2-9}\)
\(\Leftrightarrow M=\frac{2x^2}{x^2-9}\)
b) Ta có :\(P=M\left(1-N\right)\)
\(\Leftrightarrow P=\frac{2x^2}{x^2-9}\left(1-\frac{x+1}{x-3}\right)\)
\(\Leftrightarrow P=\frac{2x^2}{x^2-9}\cdot\frac{x-3-x-1}{x-3}\)
\(\Leftrightarrow P=\frac{2x^2}{x^2-9}\cdot\frac{-4}{x-3}\)
\(\Leftrightarrow P=\frac{8x^2}{x^3-3x^2-9x+27}\)
K biết có sai chỗ nào k nữa ? Check hộ mik phát
\(\)
Ta có
56 x 2 – 45 y – 40 x y + 63 x = 56 x 2 + 63 x – 45 y + 40 x y = 7 x 8 x + 9 – 5 y 8 x + 9 = 7 x - 5 y 8 x + 9
Suy ra m = 8; n = 9
Đáp án cần chọn là: A