tìm x :
a ) \(\sqrt{7^2}+3^2=x^2\)
b ) \(x^2+9=18\)
ai nhanh mk k .
K
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Những câu hỏi liên quan
Q
15 tháng 1 2018
a/ \(\left(x-1\right)\left(y+2\right)=7\)
\(\Leftrightarrow x-1;y+2\inƯ\left(7\right)\)
Suy ra :
\(\hept{\begin{cases}x-1=1\\y+2=7\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=1\\y=5\end{cases}}\)
\(\hept{\begin{cases}x-1=7\\y+2=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=8\\y=-1\end{cases}}\)
\(\hept{\begin{cases}x-1=-1\\y+2=-7\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=0\\y=-9\end{cases}}\)
\(\hept{\begin{cases}x-1=-7\\y+2=-1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-6\\x=-3\end{cases}}\)
Vậy ......
b/ \(x\left(y-3\right)=-12\)
\(\Leftrightarrow x;y-3\inƯ\left(-12\right)\)
Suy ra :
\(\hept{\begin{cases}x=1\\y-3=-12\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=1\\y=-9\end{cases}}\)
\(\hept{\begin{cases}x=-12\\y-3=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-12\\y=4\end{cases}}\)
\(\hept{\begin{cases}x=-1\\y-3=12\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=-1\\y=15\end{cases}}\)
\(\hept{\begin{cases}x=12\\y-3=-1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=12\\y=2\end{cases}}\)
Vậy ..
NT
15 tháng 1 2018
a)Ta xét: có 7 là số nguyên tố => 7= 1.7 = 7.1
\(\orbr{\begin{cases}\hept{\begin{cases}x-1=1\\y+2=7\end{cases}}\\\hept{\begin{cases}x-1=7\\y+2=1\end{cases}}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\hept{\begin{cases}x=2\\y=5\end{cases}}\\\hept{\begin{cases}x=8\\y=-1\end{cases}}\end{cases}}\)\(\orbr{\begin{cases}\hept{\begin{cases}x-1=1\\y+2=7\end{cases}}\\\hept{\begin{cases}x-1=7\\y+2=1\end{cases}}\end{cases}}\Leftrightarrow\)\(\hept{\begin{cases}x-1=7\\y+2=1\end{cases}}\)hay \(\hept{\begin{cases}x-1=1\\y+2=7\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=8\\y=-1\end{cases}}\) hay \(\hept{\begin{cases}x=2\\y=5\end{cases}}\)
b)x(y-3)=-12
Ta có: -12=1.(-12)=2.(-6)=3.(-4)=4.(-3)=(-6).2=(-12).1
Bạn xét nghiệm theo từng cặp giá trị tương ứng (12 cặp) sẽ tìm được nghiệm
c) tương tự câu b
H9
HT.Phong (9A5)
CTVHS
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left(2x+1\right)^2=6^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)
\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
25 tháng 7 2023
a) \(\sqrt{4x^2+4x+1}=6\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)
\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)
16 tháng 4 2018
tớ chỉ làm được phần a)thôi
ta có:\(\left(x-\frac{2}{3}\right):\frac{-3}{7}=\frac{-9}{11}\)
\(x-\frac{2}{3}=\frac{-9}{11}.\frac{-3}{7}\)
\(x-\frac{2}{3}=\frac{27}{77}\)
\(x=\frac{27}{77}+\frac{2}{3}\)
\(x=\frac{235}{231}\)
vậy x=\(\frac{235}{231}\)
LL
16 tháng 4 2018
a) \(\left(x-\frac{2}{3}\right)\):\(\frac{-3}{7}\)=\(\frac{-9}{1}\)
\(x-\frac{2}{3}\)=\(-9\cdot\frac{-3}{7}\)
\(x-\frac{2}{3}\)= \(\frac{27}{7}\)
\(x\)= \(\frac{27}{7}\)\(+\frac{2}{3}\)
\(x\)=\(\frac{95}{21}\)
b)\(x:2+x:\frac{5}{2}+x:\frac{1}{3}=\frac{-18}{25}\)
\(x:\left(2+\frac{5}{2}+\frac{1}{3}\right)=\frac{-18}{25}\)
\(x:\frac{12+15+2}{6}=\frac{-18}{25}\)
\(x:\frac{29}{6}=\frac{-18}{25}\)
\(x=\frac{-18}{25}\cdot\frac{29}{6}\)
\(x=\frac{-87}{25}\)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
K
24 tháng 7 2019
a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
24 tháng 7 2019
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
HK
10 tháng 3 2020
a) \(\frac{2}{5}+x=\frac{3}{4}\)
\(x=\frac{3}{4}-\frac{2}{5}\)
\(x=\frac{15}{20}-\frac{8}{20}\)
\(x=\frac{7}{20}\)
\(\)b)
\(x-\frac{1}{15}=\frac{3}{10}\\ x=\frac{3}{10}+\frac{1}{15}\\ x=\frac{9}{30}+\frac{2}{30}\\ x=\frac{11}{30}\)
c)
\(\frac{9}{8}-x=\frac{5}{12}\\ x=\frac{9}{8}-\frac{5}{12}\\ x=\frac{27}{24}-\frac{10}{24}\\ x=\frac{17}{24}\)
d)
\(\frac{3}{5}+x=\frac{5}{4}+\frac{7}{10}\\ \frac{3}{5}+x=\frac{25}{20}+\frac{14}{20}\\\frac{3}{5}+x=\frac{39}{20}\\ x=\frac{39}{20}-\frac{3}{5}\\ x=\frac{39}{20}-\frac{12}{20}\\ x=\frac{27}{20} \)
e)
\(\frac{9}{8}-x=\frac{3}{20}+\frac{2}{5}\\ \frac{9}{8}-x=\frac{3}{20}+\frac{8}{20}\\ \frac{9}{8}-x=\frac{11}{20}\\ x=\frac{9}{8}-\frac{11}{20}\\ x=\frac{45}{40}-\frac{22}{40}\\ x=\frac{23}{40}\)
g)
\(x+\frac{1}{3}=\frac{5}{6}+1\frac{7}{10}\\ x+\frac{1}{3}=\frac{5}{6}+\frac{17}{10}\\ x+\frac{1}{3}=\frac{25}{30}+\frac{51}{30}\\ x+\frac{1}{3}=\frac{76}{30}=\frac{38}{15}\\ x=\frac{38}{15}-\frac{1}{3}\\ x=\frac{38}{15}-\frac{5}{15}\\ x=\frac{33}{15}=\frac{11}{5}\)
h)
\(3x-\frac{3}{5}=\frac{1}{2}\\ 3x=\frac{1}{2}+\frac{3}{5}\\ 3x=\frac{5}{10}+\frac{6}{10}\\ 3x=\frac{11}{10}\\ x=\frac{11}{10}:3\\ x=\frac{11}{10}\cdot\frac{1}{3}\\ x=\frac{11}{30}\)
i)
\(4x+\frac{5}{12}+\frac{4}{9}=1\frac{13}{18}\\ 4x+\frac{5}{12}+\frac{4}{9}=\frac{31}{18}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{4}{9}\\ 4x+\frac{5}{12}=\frac{31}{18}-\frac{8}{18}\\ 4x+\frac{5}{12}=\frac{23}{18}\\ 4x=\frac{23}{18}-\frac{5}{12}\\ 4x=\frac{46}{36}-\frac{15}{36}\\ 4x=\frac{31}{36}\\ x=\frac{31}{36}:4\\ x=\frac{31}{36}\cdot\frac{1}{4}\\ x=\frac{31}{144}\)
k)
\(2-\left(3x+\frac{3}{7}\right)=\frac{9}{21}\\ 2-\left(3x+\frac{3}{7}\right)=\frac{3}{7}\\3x+\frac{3}{7}=2-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{14}{7}-\frac{3}{7}\\ 3x+\frac{3}{7}=\frac{11}{7}\\ 3x=\frac{11}{7}-\frac{3}{7}\\ 3x=\frac{8}{7}\\ x=\frac{8}{7}:3\\ x=\frac{8}{7}\cdot\frac{1}{3}\\ x=\frac{8}{21} \)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 4 2023
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
SH
2
15 tháng 12 2017
9-25=(7-x)-(25+7)
-16=7-x-32
7-x-32=16
7-x=-16+32
7-x=16
x=16+7
x=23
B,35-3.|x|=5.(2^3-4)
35-3.x=5.(8-4)
35-3.x=5.4
35-3.x=20
3.x=35-20
3.x=15
X=15:3
x =5
C, -6x=18
-6.x=18
X=18:(-6)
X= -3
D,10+2.|x|=2.(3^2-1)
10+2.x=2.8
10+2.x=16
2.x=16-10
2.x =6
X=6:2
x=3
Ok k mk nha
17 tháng 12 2017
a) 9 - 25 = ( 7 - x) - (25+7)
- 16 = 7-x - 32
7 - x - 32 = -16
7 - x = -16 + 32
7 - x = 16
x = 16 + 7
x = 23
b) -6x = 18
- 6 .x = 18
x = 18 : (-6)
x = -3
1 tháng 2 2017
1) 5.( x - 6 ) - 2.( x + 9 ) = 21
5x - 30 - 2x - 18 = 21
3x - 48 = 21
3x = 21 + 48
3x = 69
x = 23
2) 2.( x + 3 ) + 3.( x + 1 ) = 15 - ( - 9 )
2x + 6 + 3x + 3 = 24
5x + 9 = 24
5x = 24 - 9
5x = 15
x = 3
3) ( - x + 5 ).(3 - x ) = 0
=> - x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
4) ( x - 12 ) - 15 = ( 20 - 7 ) - ( 18 + x )
x - 12 - 15 = 13 - 18 - x
x - 27 = - 5 - x
x + x = - 5 + 27
2x = 22
x = 11
5) x - ( 17 - 8 ) = 5 + ( 10 - 3x )
x - 9 = 5 + 10 - 3x
x + 3x = 15 + 9
4x = 24
x = 6
a)
<=> 7 + 9 = x^2
<=> 16 = x^2
<=> 4 = x
b)
<=> x^2 = 18 - 9
<=> x^2 = 9
<=> x = 3
a) √72 + 32 = 7 + 9 = 16 = 42 = x2
Vậy x=4
b) x2 + 9 =18
<=> x2 = 18 -9
<=> x2 = 9 = 32
Vậy x=3