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a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
giải giúp mình bài này ới ạ mình đng cần gấp
Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
Lời giải:ĐK: $a\geq 0; a\neq 9; a\neq 4$
a)
\(A=\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{\sqrt{a}+3}{\sqrt{a}-2}+\frac{2\sqrt{a}+1}{\sqrt{a}-3}\)
\(\frac{2\sqrt{a}-9}{(\sqrt{a}-2)(\sqrt{a}-3)}-\frac{(\sqrt{a}+3)(\sqrt{a}-3)}{(\sqrt{a}-2)(\sqrt{a}-3)}+\frac{(2\sqrt{a}+1)(\ \sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}\)
\(=\frac{2\sqrt{a}-9-(a-9)+(2a-3\sqrt{a}-2)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{a-\sqrt{a}-2}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{(\sqrt{a}-2)(\sqrt{a}+1)}{(\sqrt{a}-3)(\sqrt{a}-2)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)
b) Để \(A< 1\Leftrightarrow \frac{\sqrt{a}+1}{\sqrt{a}-3}<1\Leftrightarrow 1+\frac{4}{\sqrt{a}-3}<1\)
\(\Leftrightarrow \frac{4}{\sqrt{a}-3}< 0\Leftrightarrow \sqrt{a}-3< 0\Leftrightarrow 0\leq a< 9\)
Kết hợp ĐKXĐ: suy ra $0\leq a< 9; a\neq 4$
c) Với $a$ nguyên, \(A=1+\frac{4}{\sqrt{a}-3}\in\mathbb{Z}\Leftrightarrow 4\vdots \sqrt{a}-3\)
$\Rightarrow \sqrt{a}-3\in\left\{\pm 1; \pm 2;\pm 4\right\}$
$\Rightarrow a\in\left\{4;16; 1;25; 49\right\}$
Kết hợp ĐKXĐ suy ra $a\in\left\{16;1;25;49\right\}$
ĐKXĐ: \(\left\{{}\begin{matrix}a\ge0\\a\notin\left\{4;9\right\}\end{matrix}\right.\)
a) Ta có: \(A=\dfrac{2\sqrt{a}-9}{a-5\sqrt{a}+6}-\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{2\sqrt{a}+1}{3-\sqrt{a}}\)
\(=\dfrac{\left(2\sqrt{a}-9\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}-\dfrac{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}+\dfrac{\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)
\(=\dfrac{2\sqrt{a}-9-\left(a-9\right)+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{2a-\sqrt{a}-11-a+9}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{a-2\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}-2\right)+\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}-3}\)
b) Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-1< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}+1}{\sqrt{a}-3}-\dfrac{\sqrt{a}-3}{\sqrt{a}-3}< 0\)
\(\Leftrightarrow\dfrac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\)
\(\Leftrightarrow\dfrac{4}{\sqrt{a}-3}< 0\)
mà 4>0
nên \(\sqrt{a}-3< 0\)
\(\Leftrightarrow\sqrt{a}< 3\)
hay a<9
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)
Vậy: Để A<1 thì \(\left\{{}\begin{matrix}0\le a< 9\\a\ne4\end{matrix}\right.\)
c) Để A nguyên thì \(\sqrt{a}+1⋮\sqrt{a}-3\)
\(\Leftrightarrow\sqrt{a}-3+4⋮\sqrt{a}-3\)
mà \(\sqrt{a}-3⋮\sqrt{a}-3\)
nên \(4⋮\sqrt{a}-3\)
\(\Leftrightarrow\sqrt{a}-3\inƯ\left(4\right)\)
\(\Leftrightarrow\sqrt{a}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
mà \(\sqrt{a}-3\ge-3\forall a\) thỏa mãn ĐKXĐ
nên \(\sqrt{a}-3\in\left\{1;-1;2;-2;4\right\}\)
\(\Leftrightarrow\sqrt{a}\in\left\{4;2;5;1;7\right\}\)
\(\Leftrightarrow a\in\left\{16;4;25;1;49\right\}\)
Kết hợp ĐKXĐ, ta được: \(a\in\left\{1;16;25;49\right\}\)
Vậy: Để A nguyên thì \(a\in\left\{1;16;25;49\right\}\)
a. ĐK \(\hept{\begin{cases}a\ge0\\a\ne4\\a\ne9\end{cases}}\)
P=\(\frac{2\sqrt{a}-9-\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)+\left(2\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}\)
\(=\frac{2\sqrt{a}-9-a+9+2a-4\sqrt{a}+\sqrt{a}-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}\)
\(=\frac{a-\sqrt{a}-2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}=\frac{\sqrt{a}+1}{\sqrt{a}-3}\)
b. P = \(\frac{\sqrt{a}+1}{\sqrt{a}-3}=1+\frac{4}{\sqrt{a}-3}\)
P nguyên \(\sqrt{a}-3\inƯ\left(4\right)\Rightarrow\sqrt{a}-3\in\left\{-4;-2;-1;1;2;4\right\}\)
\(\Rightarrow\sqrt{a}\in\left\{1;2;4;5;7\right\}\Rightarrow a\in\left\{1;4;16;25;49\right\}\)
c. \(P< 1\Rightarrow P-1< 0\Rightarrow\frac{\sqrt{a}+1-\sqrt{a}+3}{\sqrt{a}-3}< 0\Rightarrow\frac{4}{\sqrt{a}-3}< 0\)
\(\Rightarrow0\le a< 9\)và \(a\ne4\)
(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3}{\sqrt{x}+3}.\)
(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).
Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)
\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)
(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)
\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).
Vậy: \(x\in\varnothing.\)
(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)
\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))
\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)
(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)
Vậy: \(x=0.\)
ĐK \(\hept{\begin{cases}x\ge0\\x\ne4;x\ne9\end{cases}}\)
a. Ta có \(A=\frac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
b. Để \(A< 1\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Rightarrow\frac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\Rightarrow\frac{4}{\sqrt{x}-3}< 0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow0\le x< 9\)
Kết hợp đk thì \(0\le x< 9\)và \(x\ne4\)thì \(A< 1\)
\(\frac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\frac{\sqrt{x}+3}{x-2}-\frac{2\sqrt{x}+1}{3\sqrt{x}}\)
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)
Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)