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a: Ta có: \(\sqrt{\sqrt{x}+3}=4\)
\(\Leftrightarrow\sqrt{x}+3=16\)
\(\Leftrightarrow\sqrt{x}=13\)
hay x=169
b: Ta có: \(\sqrt{x+3}=\sqrt{1-5x}\)
\(\Leftrightarrow x+3=1-5x\)
\(\Leftrightarrow6x=-2\)
hay \(x=-\dfrac{1}{3}\left(nhận\right)\)
a) \(\sqrt{3+\sqrt{x}}=4\left(đk:x\ge0\right)\)
\(\Leftrightarrow3+\sqrt{x}=16\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\left(tm\right)\)
b) \(\sqrt{x+3}=\sqrt{1-5x}\left(đk:\dfrac{1}{5}\ge x\ge-3\right)\)
\(\Leftrightarrow x+3=1-5x\Leftrightarrow6x=-2\Leftrightarrow x=-\dfrac{1}{3}\left(ktm\right)\)
Vậy \(S=\varnothing\)
c) \(\sqrt{x^2+6x+9}=3x-1\left(đk:x\ge\dfrac{1}{3}\right)\)
\(\Leftrightarrow\sqrt{\left(x+3\right)^2}=3x-1\)
\(\Leftrightarrow\left|x+3\right|=3x-1\)
\(\Leftrightarrow x+3=3x-1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)
a: Ta có: \(2\sqrt{2}-\dfrac{1}{2}\cdot\sqrt{x}=0\)
\(\Leftrightarrow\sqrt{x}\cdot\dfrac{1}{2}=2\sqrt{2}\)
\(\Leftrightarrow\sqrt{x}=4\sqrt{2}\)
hay x=32
b: Ta có: \(2\sqrt{x}-\sqrt{\dfrac{x}{3}}=1\)
\(\Leftrightarrow2\sqrt{x}-\dfrac{\sqrt{3}}{3}\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{6+\sqrt{3}}{11}\)
hay \(x=\dfrac{39+12\sqrt{3}}{121}\)
c: Ta có: \(4\sqrt{x}+\sqrt{\dfrac{x}{2}}=\dfrac{1}{3}\)
\(\Leftrightarrow4\sqrt{x}+\dfrac{\sqrt{2}}{2}\sqrt{x}=\dfrac{1}{3}\)
\(\Leftrightarrow\sqrt{x}=\dfrac{8-\sqrt{2}}{93}\)
hay \(x=\dfrac{66-16\sqrt{2}}{8649}\)
a) \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x+3\right)\left(x-1\right)}=-\left(x+3+x-1-6\right)\)\(\left(Đk:x\ge1\right)\)
\(\left(\sqrt{x-1}+\sqrt{x+3}\right)^2+\sqrt{x-1}+\sqrt{x-3}-6=0\)
\(\left(\sqrt{x-1}+\sqrt{x+3}+3\right)\left(\sqrt{x-1}+\sqrt{x+3}-2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
b) \(\sqrt{3x-2}+\sqrt{x-1}=3x-2+x-1-6+2\sqrt{\left(3x-2\right)\left(x-1\right)}\left(Đk:x\ge1\right)\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}\right)^2-\left(\sqrt{3x-2}+\sqrt{x-1}\right)-6=0\)
\(\left(\sqrt{3x-2}+\sqrt{x-1}-3\right)\left(\sqrt{3x-2}+\sqrt{x-1}+2\right)=0\)
Đến đây em xét các trường hợp rồi bình phương lên là được nha
a/ ĐKXĐ: $x\geq 1$
Đặt $\sqrt{x-1}=a; \sqrt{x+3}=b$ thì pt trở thành:
$a+b+2ab=6-(a^2+b^2)$
$\Leftrightarrow a^2+b^2+2ab+a+b-6=0$
$\Leftrightarrow (a+b)^2+(a+b)-6=0$
$\Leftrightarrow (a+b-2)(a+b+3)=0$
Hiển nhiên do $a\geq 0; b\geq 0$ nên $a+b+3>0$. Do đó $a+b-2=0$
$\Leftrightarrow a+b=2$
Mà $b^2-a^2=(x+3)-(x-1)=4$
$\Leftrightarrow (b-a)(b+a)=4\Leftrightarrow (b-a).2=4\Leftrightarrow b-a=2$
$\Rightarrow \sqrt{x+3}=b=(a+b+b-a):2=(2+2):2=2$
$\Leftrightarrow x=1$ (tm)
\(a)ĐK:x\ge-1\\ \Leftrightarrow x+1=2\sqrt{x+1}\\ \Leftrightarrow x^2+2x+1=4x+4\\ \Leftrightarrow x^2+2x-4x+1-4=0\\ \Leftrightarrow x^2-2x-3=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{3;-1\right\}\)
\(b)ĐK:x\ge2\\ \Leftrightarrow2x-4=\sqrt{x-2}\\ \Leftrightarrow4x^2-16x+16=x-2\\ \Leftrightarrow4x^2-16x-x+16+2=0\\ \Leftrightarrow4x^2-17x+18=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{9}{4};2\right\}\)
\(c)ĐK:x\ge3\\ \Leftrightarrow2\sqrt{9\left(x-3\right)}-\dfrac{1}{5}\sqrt{25\left(x-3\right)}-\dfrac{1}{7}\sqrt{49\left(x-3\right)}=20\\ \Leftrightarrow2.3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=20\\ \Leftrightarrow6\sqrt{x-3}-\sqrt{x-3}-\sqrt{x-3}=20\\ \Leftrightarrow4\sqrt{x-3}=20\\ \Leftrightarrow\sqrt{x-3}=5\\ \Leftrightarrow x-3=25\\ \Leftrightarrow x=25+3\\ \Leftrightarrow x=28\left(tm\right)\)
Vậy \(S=\left\{28\right\}\)
a)√x−1=2(x≥1)
\(x-1=4
\)
x=5
b)
\(\sqrt{3-x}=4\) (x≤3)
\(\left(\sqrt{3-x}\right)^2=4^2\)
x-3=16
x=19
a: Ta có: \(\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)
hay x=5
b: Ta có: \(\sqrt{3-x}=4\)
\(\Leftrightarrow3-x=16\)
hay x=-13
c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)
\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)
\(\Leftrightarrow-2x=-\dfrac{47}{16}\)
hay \(x=\dfrac{47}{32}\)
d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)
\(\Leftrightarrow x-1=\dfrac{49}{4}\)
hay \(x=\dfrac{53}{4}\)
e: Ta có: \(\sqrt{x-1}-3=1\)
\(\Leftrightarrow\sqrt{x-1}=4\)
\(\Leftrightarrow x-1=16\)
hay x=17
f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)
\(\Leftrightarrow x+2=\dfrac{1}{64}\)
hay \(x=-\dfrac{127}{64}\)
\(a,3\sqrt{2x}+\sqrt{8x}-\sqrt{18x}=16\left(dk:x\ge0\right)\\ \Leftrightarrow3\sqrt{2x}+2\sqrt{2x}-3\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}\left(3+2-3\right)=16\\ \Leftrightarrow2\sqrt{2x}=16\\ \Leftrightarrow\sqrt{2x}=8\\ \Leftrightarrow\left|2x\right|=64\\ \Leftrightarrow2x=64\\ \Leftrightarrow x=32\left(tm\right)\)
Vậy \(S=\left\{32\right\}\)
\(b,\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\left(dk:x\ge-5\right)\)
\(\Leftrightarrow\sqrt{4\left(x+5\right)}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9\left(x+5\right)}=6\\ \Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}\left(2-3+4\right)=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=2\\ \Leftrightarrow\left|x+5\right|=4\\ \Leftrightarrow x+5=4\\ \Leftrightarrow x=-1\left(tm\right)\)
Vậy \(S=\left\{-1\right\}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
`a)sqrt{1-4x+4x^2}+5=x-2`
`<=>\sqrt{(2x-1)^2}=x-2-5`
`<=>|2x-1|=x-7(x>=7)`
`<=>[(2x-1=x-7),(2x-1=7-x):}`
`<=>[(x=-6(ktm)),(3x=8):}`
`<=>x=8/3(ktm)`
Vậy PTVN
`b)3sqrt{12+4x}+4/7sqrt{147+49x}=3/2sqrt{48+16x}+4(x>=-3)`
`<=>6sqrt{x+3}+4sqrt{x+3}=6sqrt{x+3}+4`
`<=>4sqrt{x+3}=4`
`<=>sqrt{x+3}=1<=>x+3=1`
`<=>x=-2(tm)`
Vậy `S={-2}`
a) \(\sqrt{1-4x+4x^2}+5=x-2\Leftrightarrow\sqrt{\left(1-2x\right)^2}+5=x-2\Leftrightarrow\left|1-2x\right|=x-7\left(1\right)\)TH1: \(1-2x\ge0\Leftrightarrow x\le\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow1-2x=x-7\Leftrightarrow3x=8\Leftrightarrow x=\dfrac{8}{3}\)(không thỏa đk)
TH2: \(1-2x< 0\Leftrightarrow x>\dfrac{1}{2}\)
\(\left(1\right)\Leftrightarrow2x-1=x-7\Leftrightarrow x=-6\)(không thỏa đk)
Vậy \(S=\varnothing\)
b) \(3\sqrt{12+4x}+\dfrac{4}{7}\sqrt{147+49x}=\dfrac{3}{2}\sqrt{48+16x}+4\Leftrightarrow6\sqrt{3+x}+4\sqrt{3+x}=6\sqrt{3+x}+4\Leftrightarrow4\sqrt{3+x}=4\Leftrightarrow\sqrt{3+x}=1\Leftrightarrow3+x=1\Leftrightarrow x=-2\)
a) Ta có: \(\sqrt{x}-1=6\)
\(\Leftrightarrow\sqrt{x}=7\)
hay x=49
Vậy: x=49
b) Ta có: \(\sqrt{x}< 4\)
nên x<16
Vậy: \(0\le x< 16\)