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Giải:
a) \(\sqrt{4x-20}-3\sqrt{\dfrac{x-5}{9}}=\sqrt{1-x}\)
\(\Leftrightarrow\sqrt{4\left(x-5\right)}-3\sqrt{x-5.\dfrac{1}{9}}=\sqrt{1-x}\)
\(\Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
\(\Leftrightarrow\sqrt{x-5}=\sqrt{1-x}\)
\(\Leftrightarrow x-5=1-x\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b) \(\sqrt{4x+8}+2\sqrt{x+2}-\sqrt{9x+18}=1\)
\(\Leftrightarrow\sqrt{4\left(x+2\right)}+2\sqrt{x+2}-\sqrt{9\left(x+2\right)}=1\)
\(\Leftrightarrow2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1\)
\(\Leftrightarrow\sqrt{x+2}=1\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)
d) \(\sqrt{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}\right)^2-7^2}=2\)
\(\Leftrightarrow\sqrt{x-49}=2\)
\(\Leftrightarrow x-49=4\)
\(\Leftrightarrow x=53\)
Vậy ...
Câu c bạn xem lại đề, mình làm không ra, kết quả xấu
Câu c nè
Đặt \(3x=a\)
=>\(9x^2=a^2\)
Đăt \(x+2=b\)
=>\(\left(x+2\right)^2=b^2\)
ta có
\(a-b=3x-x-2=2x-2\)
<=>\(2x=a-b+2\)
Khi đó pt đã cho trở thành
\(2+3\sqrt[3]{a^2b}=a-b+3\sqrt[3]{ab^2}\)\(a-b+3\sqrt[3]{ab^2}-3\sqrt[3]{a^2b}=\left(\sqrt[3]{a}\right)^3-3\sqrt[3]{a^2b}+3\sqrt[3]{ab^2}-b^3=0\)
<=>\(\left(\sqrt[3]{a}-\sqrt[3]{b}\right)^3=0\)
<=>\(\sqrt[3]{a}=\sqrt[3]{b}\)
<=>a=b
=>3x=x+2
<=>2x-2=0
<=>x=1
nhớ tick nha
b)\(\sqrt{4x-8}+2\sqrt{9x-18}-\sqrt{x-2}=14\)
Đk:\(x\ge2\)
\(pt\Leftrightarrow\sqrt{4x-8}-4+2\sqrt{9x-18}-12-\left(\sqrt{x-2}-2\right)=0\)
\(\Leftrightarrow\frac{4x-8-16}{\sqrt{4x-8}+4}+\frac{4\left(9x-18\right)-144}{2\sqrt{9x-18}+12}-\frac{x-2-4}{\sqrt{x-2}+2}=0\)
\(\Leftrightarrow\frac{4\left(x-6\right)}{\sqrt{4x-8}+4}+\frac{36\left(x-6\right)}{2\sqrt{9x-18}+12}-\frac{x-6}{\sqrt{x-2}+2}=0\)
\(\Leftrightarrow\left(x-6\right)\left(\frac{4}{\sqrt{4x-8}+4}+\frac{36}{2\sqrt{9x-18}+12}-\frac{1}{\sqrt{x-2}+2}\right)=0\)
Thấy: \(\frac{4}{\sqrt{4x-8}+4}+\frac{36}{2\sqrt{9x-18}+12}-\frac{1}{\sqrt{x-2}+2}=0\) vô nghiệm
Nên x-6=0 suy ra x=6
Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
1: =>|2x-1|=5
=>2x-1=5 hoặc 2x-1=-5
=>2x=6 hoặc 2x=-4
=>x=3 hoặc x=-2
2: \(\Leftrightarrow2\sqrt{x-3}+\dfrac{1}{3}\cdot3\sqrt{x-3}-\sqrt{x-3}=4\)
\(\Leftrightarrow\sqrt{x-3}=2\)
=>x-3=4
hay x=7
5: \(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=>x-2=0 hoặc x+2=1
=>x=2 hoặc x=-1
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3