giải giúp mình với ạ, đây là bài rút gọn biểu thức nha
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a) \(\sqrt{0,64.a^2}\left(a>0\right)=0,8.\left|a\right|=0,8a\)
b) \(\sqrt{a^2\left(a-2\right)^2}\left(a>2\right)=\left|a\left(a-2\right)\right|=a\left(a-2\right)=a^2-2a\)
c) \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=9x^4-\left(2x+1\right)^2-\left(9x^4+6x^2+1\right)\\ =9x^4-4x^2-4x-1-9x^4-6x^2-1\\ =-10x^2-4x-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\left(\dfrac{\sqrt{x}}{2x}-\dfrac{1}{\sqrt{x}}\right)\cdot\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\\ =\left(\dfrac{\sqrt{x}}{2x}-\dfrac{2\sqrt{x}}{2x}\right)\cdot\left(\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\\ =\dfrac{x-2\sqrt{x}}{2x}\cdot\dfrac{x-2\sqrt{x}+1-\left(x+2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2x}\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{2x}\cdot\dfrac{-4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(\sqrt{x}-2\right)}{x-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2sin8a-sin16a}{2sin8a+sin16a}=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(1-2sin^24a\right)}=\dfrac{2sin^24a}{2-2sin^24a}=\dfrac{sin^24a}{1-sin^24a}=\dfrac{sin^24a}{cot^24a}=tan^24a\)
\(=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}\)
\(=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(2cos^24a-1\right)}=\dfrac{2sin^24a}{2cos^24a}=tan^24a\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
a. \(\sqrt{x+2}\) có nghĩa khi \(x+2\ge0\Leftrightarrow x\ge-2\)
Vậy biểu thức \(\sqrt{x+2}\) có nghĩa khi \(x\ge-2\)
b. \(\left\{{}\begin{matrix}2x+y=5\\x+2y=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x+y=5\\2x+4y=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3y=3\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất (x; y) = (2; 1)
c. \(A=\left(\dfrac{3}{x-3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right).\dfrac{x-9}{\sqrt{x}}\left(x>0;x\ne9\right)\)
\(=\left[\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}\left(x-9\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}\left(x-9\right)}\right].\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{3\sqrt{x}+9+x-3\sqrt{x}}{\sqrt{x}\left(x-9\right)}.\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{x+9}{\sqrt{x}\left(x-9\right)}.\dfrac{x-9}{\sqrt{x}}\)
\(=\dfrac{x+9}{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
13.
$(x+4)^2+(x+5)(x-5)-2x(x+1)$
$=(x^2+8x+16)+(x^2-25)-(2x^2+2x)$
$=x^2+8x+16+x^2-25-2x^2-2x$
$=(x^2+x^2-2x^2)+(8x-2x)+(16-25)=6x-9$
14.
$(x-1)^2-2(x+3)(x-3)+4x(x-4)$
$=(x^2-2x+1)-2(x^2-9)+(4x^2-16x)$
$=x^2-2x+1-2x^2+18+4x^2-16x$
$=(x^2-2x^2+4x^2)+(-2x-16x)+(1+18)=3x^2-18x+19$
15.
$(y-3)(y+3)(y^2+9)-(y^2+2)(y^2-2)$
$=(y^2-9)(y^2+9)-(y^4-4)$
$=(y^4-81)-(y^4-4)=-81+4=-77$
![](https://rs.olm.vn/images/avt/0.png?1311)
a,hđt số 3 = \(\left(a^2+2a\right)^2-9\)
b,hđt số 3=\(\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)(đổi dấu làm ngoặc khi trước nó là dấu trừ)=\(x^2-\left(y-6\right)^2\)
a) \(\left(a^2+2a+3\right)\left(a^2+2a-3\right)\)
\(=\left(a^2+2a\right)^2+3.\left(-3\right)\)
\(=\left(a^2+2a\right)^2-9\)
b) \(\left(x-y+6\right)\left(x+y-6\right)\)
\(=\left[x-\left(y-6\right)\right]\left[x+\left(y-6\right)\right]\)
\(=x^2-\left(y-6\right)^2\)
d: \(\dfrac{-\left(\sqrt{3}-\sqrt{6}\right)}{1-\sqrt{2}}+\dfrac{6\sqrt{3}+3}{\sqrt{3}}-\dfrac{13}{4+\sqrt{3}}\)
\(=-\sqrt{3}+6+\sqrt{3}-4+\sqrt{3}\)
\(=2+\sqrt{3}\)