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Tớ làm nốt nè :3
\(1b.3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}=8\sqrt{2}\)
\(c.\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=4\)
\(2a.\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow4x^2-4x+1=9\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(b.\sqrt{4x-4}-\sqrt{9x-9}+5\sqrt{x-1}=7\left(x\ge1\right)\)
\(\Leftrightarrow2\sqrt{x-1}-3\sqrt{x-1}+5\sqrt{x-1}=7\)
\(\Leftrightarrow4\sqrt{x-1}=7\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{4}\)
\(\Leftrightarrow x=\dfrac{65}{16}\)
c. Sai đề.
Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
Mình làm một vài câu thôi nhé, các câu còn lại tương tự.
Giải:
a) ??? Đề thiếu
b) \(\sqrt{-3x+4}=12\)
\(\Leftrightarrow-3x+4=144\)
\(\Leftrightarrow-3x=140\)
\(\Leftrightarrow x=\dfrac{-140}{3}\)
Vậy ...
c), d), g), h), i), p), q), v), a') Tương tự b)
w), x) Mình đã làm ở đây:
Câu hỏi của Ami Yên - Toán lớp 9 | Học trực tuyến
z) \(\sqrt{16\left(x+1\right)^2}-\sqrt{9\left(x+1\right)^2}=4\)
\(\Leftrightarrow4\left(x+1\right)-3\left(x+1\right)=4\)
\(\Leftrightarrow x+1=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b') \(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+1}\)
\(\Leftrightarrow3\sqrt{x+1}+2\sqrt{x+1}-\sqrt{x+1}=0\)
\(\Leftrightarrow4\sqrt{x+1}=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...
- Câu a có chút thiếu sót, mong thông cảm :)
\(\sqrt{3x-1}\) = 4
Lời giải:
Bạn cứ nhớ công thức $\sqrt{x^2}=|x|$, rồi dùng điều kiện đề bài để phá dấu trị tuyệt đối là được
a)
\(\sqrt{16a^2}-5a=\sqrt{(4a)^2}-5a=|4a|-5a=4a-5a=-a\)
b)
\(3x+2-\sqrt{9x^2+6x+1}=3x+2-\sqrt{(3x)^2+2.3x.1+1^2}\)
\(=3x+2-\sqrt{(3x+1)^2}=3x+2-|3x+1|=3x+2-(3x+1)=1\)
c)
\(\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{7+1+2.\sqrt{7}.\sqrt{1}}-\sqrt{7}\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{7}=|\sqrt{7}+1|-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
d)
\(\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{13+1-2\sqrt{13}}+\sqrt{13+1+2\sqrt{13}}\)
\(=\sqrt{(\sqrt{13}-1)^2}+\sqrt{(\sqrt{13}+1)^2}=|\sqrt{13}-1|+|\sqrt{13}+1|\)
\(=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)
e)
\(2x-\sqrt{4x^2-4x+1}=2x-\sqrt{(2x-1)^2}=2x-|2x-1|=2x-(2x-1)=1\)
g)
\(|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=|x-2|+\frac{\sqrt{(x-2)^2}}{x-2}=|x-2|+\frac{|x-2|}{x-2}\)
\(=(x-2)+\frac{(x-2)}{x-2}=x-2+1=x-1\)
\(A=\left(x-2\right)\cdot\sqrt{\dfrac{9}{\left(x-2\right)^2}}+3=\dfrac{3\left(x-2\right)}{\left|x-2\right|}+3=\dfrac{3\left(x-2\right)}{-\left(x-2\right)}=-3+3=0\)
\(B=\sqrt{\dfrac{a}{6}}+\sqrt{\dfrac{2a}{3}}+\sqrt{\dfrac{3a}{2}}=\dfrac{\sqrt{a}}{\sqrt{6}}+\dfrac{\sqrt{2a}}{\sqrt{3}}+\dfrac{\sqrt{3a}}{\sqrt{2}}=\dfrac{\sqrt{a}+2\sqrt{a}+3\sqrt{a}}{\sqrt{6}}=\dfrac{6\sqrt{a}}{\sqrt{6}}=\sqrt{6a}\)
\(E=\sqrt{9a^2}+\sqrt{4a^2}+\sqrt{\left(1-a\right)^2}+\sqrt{16a^2}=3\left|a\right|+2\left|a\right|+\left|1-a\right|+4\left|a\right|=9\left|a\right|+1-a=-9a+1-a=-10a+1\)
\(F=\left|x-2\right|\cdot\dfrac{\sqrt{x^2}}{x}=\left|x-2\right|\cdot\dfrac{\left|x\right|}{x}=\dfrac{x\left(x-2\right)}{x}=x-2\)
\(H=\dfrac{x^2+2\sqrt{3}\cdot x+3}{x^2-3}=\dfrac{\left(x+\sqrt{3}\right)^2}{\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)}=\dfrac{x+\sqrt{3}}{x-\sqrt{3}}\)
\(I=\left|x-\sqrt{\left(x-1\right)^2}\right|-2x=\left|x-\left(-\left(x-1\right)\right)\right|-2x=\left|x+x-1\right|-2x=\left|2x-1\right|-2x=1-4x\)
Làm nốt ::v
\(2.3\sqrt{\left(a-2\right)^2}=3\text{ |}a-2\text{ |}=3\left(a-2\right)\left(a< 2\right)\)
\(3.\sqrt{81a^4}+3a^2=\sqrt{3^4.a^4}+3a^2=9a^2+3a^2=12a^2\)
\(4.\sqrt{64a^2}+2a=\text{ |}8a\text{ |}+2a=8a+2a=10a\left(a>=0\right)\)
\(6.\sqrt{a^2+6a+9}+\sqrt{a^2-6a+9}=\sqrt{\left(a+3\right)^2}+\sqrt{\left(a-3\right)^2}=\text{ |}a+3\text{ |}+\text{ |}a-3\text{ |}\)
\(7.\dfrac{\sqrt{1-2x+x^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\text{ |}x-1\text{ |}}{x-1}\)
\(8.\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\dfrac{\text{ |}3x-1\text{ |}}{\left(3x-1\right)\left(3x+1\right)}\)
\(9.4-x-\sqrt{4-4x+x^2}=4-x-\sqrt{\left(x-2\right)^2}=4-x-\text{ |}x-2\text{ |}\)
Mình làm ba câu mẫu, bạn theo đó mà làm các câu còn lại.
Giải:
1) \(2\sqrt{a^2}\)
\(=2\left|a\right|\)
\(=2a\left(a\ge0\right)\)
Vậy ...
5) \(3\sqrt{9a^6}-6a^3\)
\(=3\sqrt{\left(3a^3\right)^2}-6a^3\)
\(=3.3a^3-6a^3\)
\(=9a^3-6a^3\)
\(=3a^3\)
Vậy ...
10) \(C=\sqrt{4x^2-4x+1}-\sqrt{4x^2+4x+1}\)
\(\Leftrightarrow C=\sqrt{\left(2x-1\right)^2}-\sqrt{\left(2x+1\right)^2}\)
\(\Leftrightarrow C=2x-1^2-\left(2x+1^2\right)\)
\(\Leftrightarrow C=2x-1-2x-1\)
\(\Leftrightarrow C=-2\)
Vậy ...