1. Tìm x.
a. 72x+2 = 49
b. 4x+1 = 64
c. 9x+1 = 3*16*
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QB
4
TT
29 tháng 12 2017
a,\(9x^2-72x=0\)
\(9x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}9x=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
Vậy...
NT
29 tháng 12 2017
a
\(9x^2-72x=0\\ \Leftrightarrow9x\left(x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
b)
\(\left(16-4x\right)\left(x+3\right)-\left(x+1\right)\left(4-4x\right)=40\\ \Leftrightarrow\left(16x+48-4x^2-12x\right)-\left(4x-4x^2+4-4x\right)=40\\ \Leftrightarrow4x+44=40\\ \Leftrightarrow x=-1\)
HN
2
DP
5 tháng 10 2017
a, \(x^3-5x=0\)
\(\Rightarrow x\left(x^2-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\end{matrix}\right.\)
b, \(4x^3-9x=0\)
\(\Rightarrow x\left(4x^2-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\4x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{\dfrac{9}{4}}\end{matrix}\right.\)
c, \(2x^3-72x=0\)
\(\Rightarrow2x\left(x^2-36\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm6\end{matrix}\right.\)
d, \(4\left(3x+1\right)^2+16=25\)
\(\Rightarrow4\left(3x+1\right)^2-9=0\)
\(\Rightarrow\left[2\left(3x+1\right)-3\right]\left[2\left(3x+1\right)+3\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}2\left(3x+1\right)-3=0\\2\left(3x+1\right)+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{3}{2}\\3x+1=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)
S
5 tháng 10 2017
a, \(x^2-5x=0\)
\(\Rightarrow x\left(x^2-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\end{matrix}\right.\)
b, \(4x^3-9x=0\)
\(\Rightarrow x\left(4x^2-9\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\4x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{9}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{\dfrac{9}{4}}\end{matrix}\right.\)
c, \(2x^3-72x=0\)
\(\Rightarrow2x\left(x^2-36\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x^2-36=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
AH
Akai Haruma
Giáo viên
16 tháng 7 2020
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow-2\sqrt{x-1}=-2\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{2}{2}\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\left(tm\right)\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
15 tháng 10 2021
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
HT
2
11 tháng 8 2021
a: Ta có: \(\sqrt{x}< 3\)
nên \(0\le x< 9\)
b: Ta có: \(\sqrt{4x+16}+\sqrt{x+4}+2\sqrt{9x+36}=35\)
\(\Leftrightarrow2\sqrt{x+4}+\sqrt{x+4}+6\sqrt{x+4}=35\)
\(\Leftrightarrow\sqrt{x+4}=\dfrac{35}{9}\)
\(\Leftrightarrow x+4=\dfrac{1225}{81}\)
hay \(x=\dfrac{901}{81}\)
NT
11 tháng 8 2021
a) \(\sqrt{x}< 3\Rightarrow x< 9\)
b) \(\sqrt{4x+16}+\sqrt{x+4}+2\sqrt{9x+36}=35\)
\(\Rightarrow2\sqrt{x+4}+\sqrt{x+4}+6\sqrt{x+4}=35\)
\(\Rightarrow\sqrt{x+4}=\dfrac{35}{9}\)
\(\Rightarrow x+4=\dfrac{1225}{81}\)
\(\Rightarrow x=\dfrac{901}{81}\)
c) \(\sqrt{x+2\sqrt{x-1}}=3\)
\(\Rightarrow\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=3\)
\(\Rightarrow\sqrt{\left(x-1+1\right)^2}=3\)
\(\Rightarrow\sqrt{x^2}=3\)
\(\Rightarrow\left|x\right|=3\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
TN
2
28 tháng 10 2021
\(ĐK:x\ge1\\ PT\Leftrightarrow12\sqrt{x-1}-\sqrt{x-1}-8\sqrt{x-1}+\sqrt{x-1}=16\\ \Leftrightarrow4\sqrt{x-1}=16\\ \Leftrightarrow\sqrt{x-1}=4\\ \Leftrightarrow x-1=16\\ \Leftrightarrow x=17\left(tm\right)\)
LL
1
29 tháng 6 2021
`2sqrt{36x-36}-1/3sqrt{9x-9}-4sqrt{4x-4}+sqrt{x-1}=16`
`ĐK:x>=1`
`pt<=>2sqrt{36(x-1)}-1/3sqrt{9(x-1)}-4sqrt{4(x-1)}+sqrt{x-1}=16`
`<=>12sqrt{x-1}-sqrt{x-1}-8sqrt{x-1}+sqrt{x-1}=16`
`<=>4sqrt{x-1}=16`
`<=>sqrt{x-1}=4`
`<=>x-1=16`
`<=>x=17(tmđk)`
Vậy `S={17}`
22 tháng 7 2023
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
22 tháng 7 2023
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
a,
=> 7^2x+2 = 7^2
=> 2x+2 = 2
=>2x=0
=> x=0
Vậy x=0
b, 4^x+1 = 64
=> 4^x+1 = 4^3
=> x+1=3
=> x=2
vậy x=2
c, 9^x+1= 3*16
P/S: DẤU SAO ĐÓ LÀ GÌ VẬY BẠN
a.7^2x+2=7^2
=>2x+2=2
2x=2+2
2x=4
x=4/2
x=2
b.4^x+1=64
4^x+1=4^3
=>x+1=3
x=3-1
x=2
câu c mk ko hiểu, nhưng mk giúp bn 2 câu rùi đó, tk nha+_+