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\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)
=> \(\sqrt{x^2-25}=\sqrt{x-5}\)
=>\(x^2-25=x-5\)
=>\(x^2-x=25-5=20\)
=>( đến đoạn này mình xin chịu )
\(a,\sqrt{16x}=8\)
=>\(16x=8^2\)
=>\(16x=64\)
=>\(x=64:16=4\)
Vậy \(x\in\left\{4\right\}\)
\(b,\sqrt{x^2}=2x-1\)
=>\(x=2x-1\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy \(x\in\left\{1\right\}\)
\(c,\sqrt{9.\left(x-1\right)}=21\)
=>\(9.\left(x-1\right)=21^2=441\)
=> \(x-1=441:9=49\)
=>\(x=49+1=50\)
Vậy \(x\in\left\{50\right\}\)
\(d,\sqrt{4\left(1-x\right)^2}-6=0\)
=>\(\sqrt{4\left(1-x\right)^2}=0+6=6\)
=> \(4\left(1-x\right)^2=6^2=36\)
=>\(\left(1-x\right)^2=36:4=9\)
=>\(1-x=\sqrt{9}=3\)
=>\(x=1-3=-2\)
Vậy \(x\in\left\{-2\right\}\)
\(g,\sqrt{9\left(2-3x\right)^2}=6\)
=> \(9.\left(2-3x\right)^2=6^2=36\)
=> \(\left(2-3x\right)^2=36:9=4\)
=> \(2-3x=\sqrt{4}=2\)
=>\(3x=2-2=0\)
=>\(x=0:3=0\)
Vậy \(x\in\left\{0\right\}\)
( còn các bài còn lại mình sẽ nghĩ tiếp , HS6-7 làm bài )
Bài 1 :
a) \(\sqrt{4\left(a-3\right)^2}+2\sqrt{\left(a^2+4a+4\right)}\)
= \(2\left|a-3\right|+2\left|a+2\right|\)
\(=2.\left(-a+3\right)+2\left(-a-2\right)\)
b) có sai đề ko ?
c) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}=4x-\sqrt{8}+\sqrt{\dfrac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt{4}+x=3x-2\sqrt{4}\)
a) Đề có lẽ là:
đk: \(x\ge0\)
\(\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}+2\right)x=x\sqrt{x}-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}+1+x\sqrt{x}+2x-x\sqrt{x}+\sqrt{x}-3=0\)
\(\Leftrightarrow3x+3\sqrt{x}-2=0\)
\(\Leftrightarrow3\left(x+\sqrt{x}+\frac{1}{4}\right)-\frac{11}{4}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{11}{12}=0\)
\(\Leftrightarrow\left(\sqrt{x}+\frac{3+\sqrt{33}}{6}\right)\left(\sqrt{x}+\frac{3-\sqrt{33}}{6}\right)=0\)
Vì \(\sqrt{x}\ge0\left(\forall x\right)\)
=> \(\sqrt{x}=\frac{3-\sqrt{33}}{6}\Rightarrow x=\frac{7-\sqrt{33}}{6}\)
b) đk: \(x\ge1\)
Ta có: \(\sqrt{4\left(x^2-1\right)}-2\sqrt{15}=0\)
\(\Leftrightarrow\sqrt{x^2-1}=\sqrt{15}\)
\(\Leftrightarrow x^2-1=15\)
\(\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
a)
\(\dfrac{\left(\sqrt{x^2+4}-2\right)\left(\sqrt{x^2+4}-2\right)\left(x+\sqrt{x}+1\right)\sqrt{x-2\sqrt{x}+1}}{x\left(x\sqrt{x}-1\right)}\\=\dfrac{\left(\left(\sqrt{x^2+4}\right)^2-4\right)\left(\left(x+\sqrt{x}+1\right)\sqrt{\left(x-1\right)^2}\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{\left(x^2+4-4\right)\left(\left(x+\sqrt{x}+1\right)\left(x-1\right)\right)}{x\left(x\sqrt{x}-1\right)}\\ =\dfrac{x^2\left(x^3-1\right)}{x\left(x\sqrt{x}-1\right)}=x^2\sqrt{x}\)
b)
\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}-\dfrac{\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}\right)\left(\dfrac{a}{\sqrt{a}}-\dfrac{4}{\sqrt{a}}\right)\\ =\left(\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{a-4}\right)\left(\dfrac{a-4}{\sqrt{a}}\right)\\ =\dfrac{-8\sqrt{a}}{a-4}\cdot\dfrac{a-4}{\sqrt{a}}=-8\)
c)
\(\left(\dfrac{\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)}+\dfrac{\left(\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)}\right)\left(1-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\dfrac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}}{\sqrt{a}}-\dfrac{1}{\sqrt{a}}\right)\\ =\left(\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\\ =\dfrac{2a+2}{a-1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(a+1\right)}{a+1}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ =\dfrac{-2\left(\sqrt{a}-1\right)}{\sqrt{a}}\)
d)
\(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}^3-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+x+1\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\\ =\sqrt{x}\left(\sqrt{x}-1\right)-\sqrt{x}\left(\sqrt{x}+1\right)+x+1\\ =x-\sqrt{x}-x-\sqrt{x}+x+1\\ =x-2\sqrt{x}+1\\ =\left(x-1\right)^2\)
a) \(\sqrt{\left(x-2\right)^2}=\sqrt{x-2}\)
\(\Leftrightarrow\left|x-2\right|=\sqrt{x-2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{x-2}\\-x+2=\sqrt{x-2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
Vậy ....
Mk chỉ làm được câu a thôi mong bạn thông cảm
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
\(\sqrt{4\left(x+1\right)}=\sqrt{8}\)
⇒4(x+1)=8
⇒x+1=2
⇒x=1
a. \(\sqrt{4\left(x+1\right)}=\sqrt{8}\) ĐKXĐ: \(x\ge-1\)
<=> \(\left(\sqrt{4\left(x+1\right)}\right)^2=\left(\sqrt{8}\right)^2\)
<=> 4(x + 1) = 8
<=> 4x + 4 = 8
<=> 4x = -4
<=> x = -1 (TM)
Vậy nghiệm của PT là S = \(\left\{-1\right\}\)