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a) ta có : \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3}+\sqrt{\left(\sqrt{2}+2\right)^3}=1-\sqrt{2}+\sqrt{2}+2=3\)
b) ở đây : https://hoc24.vn/hoi-dap/question/650070.html
a, Ta có: \(A+B=\sqrt[3]{7-5\sqrt{2}}+\sqrt[3]{20+14\sqrt{2}}\)
\(=\sqrt[3]{\left(1-\sqrt{2}\right)^3+\sqrt[3]{\left(\sqrt{2}+2\right)^3}}\)
\(=1-\sqrt{2}+\sqrt{2}+2=1+2=3\)
Vậy ...
\(b,x=\sqrt{6-3\sqrt{2+\sqrt{3}}}-\sqrt{2+\sqrt{2+\sqrt{3}}}\)
Đặt \(\sqrt{2+\sqrt{3}=t}\) , ta có:
\(x=\sqrt{6-3.t}-\sqrt{2+t}\)
\(\Rightarrow x^2=2+t+3.\left(2-t\right)-2\sqrt{3}\left(2+t\right)\left(2-t\right)\)
\(=8-2t-2\sqrt{3\left(4-t^2\right)}\)
\(=8-2t-2\sqrt{3\left(4-2-\sqrt{3}\right)}\)
\(=8-2t-\sqrt{6}.\sqrt{4-2\sqrt{3}}\)
\(=8-2\sqrt{2+\sqrt{3}}-\sqrt{6}\left(\sqrt{3}-1\right)\)
\(=8-\sqrt{2}.\sqrt{4+2\sqrt{3}}-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{2}\left(\sqrt{3}+1\right)-3\sqrt{2}+\sqrt{6}\)
\(=8-\sqrt{6}-\sqrt{2}-3\sqrt{2}+\sqrt{6}\)
\(=8-4\sqrt{2}\)
\(\Rightarrow x^2-8=-4\sqrt{2}\)
\(\Rightarrow\left(x^2-8\right)^2=32\)
\(\Rightarrow x^4-16x^2+64=32\)
\(\Rightarrow x^4-16x^2+64-32=0\)
\(\Rightarrow x^4-16x^2+32=0\) (đpcm)
Chúc bạn hok tốt!!!
b,
+ Với \(x=0\) \(\Rightarrow PTVN\)
+ Với \(x\ne0\), chia cả 2 vế cho \(x^2\) :
\(PT\Leftrightarrow x^2-16x+46+\frac{144}{x}+\frac{81}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\frac{81}{x^2}\right)-16\left(x-\frac{9}{x}\right)+46=0\)
Đặt \(x-\frac{9}{x}=t\Rightarrow t^2=x^2+\frac{81}{x^2}-18\)
\(\Leftrightarrow t^2+18-16t+46=0\)
\(\Leftrightarrow t^2-16t+64=0\Rightarrow t=8\)
\(\Leftrightarrow x-\frac{9}{x}=8\Leftrightarrow x^2-8x-9=0\) \(\Rightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\) (t/m)
cậu xem làm được mấy bài kia không làm giùm với (đang gấp) :))
a.
\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)
\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)
b.
\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)
\(\Leftrightarrow x^2-8=5x+1\)
\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)
............................
tương tự ..
c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)
=>x-5=0 hoặc x+5=1
=>x=-4 hoặc x=5
d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=7/2 hoặc x=-3/2
e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc 3 căn x+2=1
=>x=2 hoặc x+2=1/9
=>x=-17/9 hoặc x=2
a, \(\sqrt{x+2}-3\sqrt{x^2-4}\) = 0
⇔\(\sqrt{x+2}\) = \(3\sqrt{\left(x-2\right)\left(x+2\right)}\)
⇔\(3\sqrt{x-2}\) = 0
⇔\(\sqrt{x-2}\) = 0
⇔ x - 2 = 0
⇔ x = 2
b, \(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\)
⇔\(\sqrt{1-x}+\sqrt{4\left(1-x\right)}-\dfrac{1}{3}\sqrt{16\left(1-x\right)}+5=0\)
⇔\(\sqrt{1-x}+2\sqrt{\left(1-x\right)}-\dfrac{4}{3}\sqrt{\left(1-x\right)}+5=0\)
⇔\(\left(1+2-\dfrac{4}{3}\right)\sqrt{1-x}=-5\)
⇔\(\dfrac{5}{3}\sqrt{1-x}=-5\)
⇔\(\sqrt{1-x}=-3\) ( vô lí )
⇒ Phương trình vô nghiệm
a) \(ĐKXĐ:\left[{}\begin{matrix}x\ge2\\x\le-2\end{matrix}\right.\)
\(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
\(\Leftrightarrow\sqrt{x-2}=3\sqrt{x^2-4}\)
\(\Leftrightarrow x-2=9\left(x^2-4\right)\)
\(\Leftrightarrow x-2=9x^2-36\)
\(\Leftrightarrow9x^2-x-34=0\)
\(\Leftrightarrow9x^2-18x+17x-34=0\)
\(\Leftrightarrow9x\left(x-2\right)+17\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(9x+17\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\9x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-\dfrac{17}{9}\left(Ktm\right)\end{matrix}\right.\)
Vây: x = 2
b)\(ĐKXĐ:x\le1\)
\(\sqrt{1-x}+\sqrt{4-4x}-\dfrac{1}{3}\sqrt{16-16x}+5=0\)
\(\Leftrightarrow\sqrt{1-x}+\sqrt{4\left(1-x\right)}-\dfrac{1}{3}\sqrt{16\left(1-x\right)}+5=0\)
\(\Leftrightarrow\sqrt{1-x}+2\sqrt{\left(1-x\right)}-\dfrac{4}{3}\sqrt{\left(1-x\right)}+5=0\)
\(\Leftrightarrow\sqrt{1-x}\left(1+2-\dfrac{4}{3}\right)+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{1-x}+5=0\)
\(\Leftrightarrow\dfrac{5}{3}\sqrt{x-1}=-5\)
\(\Leftrightarrow\sqrt{1-x}=-3\left(vn\right)\)
Vậy: \(x=\varnothing\)
Sai thì thôi nhâ
4.a)\(x-2\sqrt{x}+3\)
\(=x-2\sqrt{x}+1+2\)
\(=\left(\sqrt{x}-1\right)^2+2\)
Vì \(\left(\sqrt{x}-1\right)^2\ge0,\forall x\)
\(\left(\sqrt{x}-1\right)^2+2\ge2\)
\(\Rightarrow Min_{bt}=2\) khi \(\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)
b)Ta có:
\(x-4\sqrt{y}+13\ge0\)
\(\Leftrightarrow x-4\sqrt{y}\ge-13\)
Dấu "=" xảy ra khi \(x-4\sqrt{y}=0\Leftrightarrow x=4\sqrt{y}\)
Vậy \(min_{bt}=0\) khi \(x=4\sqrt{y}\)
c)Ta có:
\(2x-4\sqrt{y}+6\ge0\)
\(\Leftrightarrow x-2\sqrt{y}+3\ge0\)
\(\Leftrightarrow x-2\sqrt{y}\ge-3\)
Dấu "=" xảy ra khi \(x-2\sqrt{y}=0\Leftrightarrow x=2\sqrt{y}\)
Vậy \(Min_{bt}=0\) khi \(x=2\sqrt{y}\)
d)Ta có:
\(x^2+2x+5=x^2+2x+1+4=\left(x+1\right)^2+4\)
Vì \(\left(x+1\right)^2\ge0,\forall x\)
\(\Leftrightarrow\left(x+1\right)^2+4\ge4\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)^2+4}\le\frac{1}{4}\)
\(\Leftrightarrow-\frac{1}{\left(x+1\right)^2+4}\ge-\frac{1}{4}\)
\(\Leftrightarrow-\frac{4}{\left(x+1\right)^2+4}\ge-1\)
Vậy \(Min_{bt}=-1\) khi \(x+1=0\Leftrightarrow x=-1\)
a)
pt <=> \(x^2=324\)
<=> \(\orbr{\begin{cases}x=18\\x=-18\end{cases}}\)
Vậy tập hợp nghiệm của pt là: S={18; -18}
b) pt <=> \(16x^2=5\)
<=> \(x^2=\frac{5}{16}\)
<=> \(\orbr{\begin{cases}x=\frac{\sqrt{5}}{4}\\x=-\frac{\sqrt{5}}{4}\end{cases}}\)
a. \(-x^2+324=0\)
\(\Leftrightarrow-x^2=-324\)
\(\Leftrightarrow x^2=324=18^2\)
\(\Leftrightarrow x=18;x=-18\)
b. \(16x^2-5=0\)
\(\Leftrightarrow16x^2=5\)
\(\Leftrightarrow x^2=\frac{5}{16}=\frac{\sqrt{5}}{4}^2\)
\(\Leftrightarrow x=\frac{\sqrt{5}}{4}\)
Bài 3: \(3\left(\sqrt{2x^2+1}-1\right)=x\left(1+3x+8\sqrt{2x^2+1}\right)\)
\(\Leftrightarrow\left(3-8x\right)\sqrt{2x^2+1}=3x^2+x+3\)
\(\Rightarrow\left(3-8x\right)^2\left(2x^2+1\right)=\left(3x^2+x+3\right)^2\)
\(\Leftrightarrow119x^4-102x^3+63x^2-54x=0\)
\(\Leftrightarrow x\left(7x-6\right)\left(17x^2+9\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{6}{7}\end{cases}}\)
Thử lại, ta nhận được \(x=0\)là nghiệm duy nhất của phương trình
\(f,\sqrt{x^2-25}-\sqrt{x-5}=0\)
=> \(\sqrt{x^2-25}=\sqrt{x-5}\)
=>\(x^2-25=x-5\)
=>\(x^2-x=25-5=20\)
=>( 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 )
\(x^3=\left(\sqrt{5}+\sqrt{3}\right)^3=\sqrt{5^3}+3.5.\sqrt{3}+3.\sqrt{5}.3+\sqrt{3^3}\)
\(=14\sqrt{5}+18\sqrt{3}\)
\(x^2=\left(\sqrt{5}+\sqrt{3}\right)^2=8+2\sqrt{15}\)
\(\frac{1}{x^2}=\frac{1}{8+2\sqrt{15}}=\frac{8-2\sqrt{15}}{8^2-4.15}=\frac{4-\sqrt{15}}{2}\)
\(\Leftrightarrow\frac{4}{x^2}=8-2\sqrt{15}\)
\(\Leftrightarrow x^2+\frac{4}{x^2}=16\)
\(\Leftrightarrow x^4-16x^2+4=0\)
Ta có đpcm.