11/3 + | x |= 9/4
giúp m với mn ơi
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Khách
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Những câu hỏi liên quan
31 tháng 12 2023
a: \(\sqrt{x^2+6x+9}=\sqrt{11+6\sqrt{2}}\)
=>\(\sqrt{\left(x+3\right)^2}=\sqrt{\left(3+\sqrt{2}\right)^2}\)
=>\(\left|x+3\right|=\left|3+\sqrt{2}\right|=3+\sqrt{2}\)
=>\(\left[{}\begin{matrix}x+3=3+\sqrt{2}\\x+3=-3-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-6-\sqrt{2}\end{matrix}\right.\)
b: \(\left\{{}\begin{matrix}2x-y=4\\x+2y=-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x-2y=8\\x+2y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x-2y+x+2y=8-3\\2x-y=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}5x=5\\y=2x-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\cdot1-4=-2\end{matrix}\right.\)
BT
1
17 tháng 9 2023
\(\dfrac{1}{3}\sqrt[]{9x+9}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4x+4}{25}}=11\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt[]{9\left(x+1\right)}-2\sqrt[]{x+1}+8\sqrt[]{\dfrac{4\left(x+1\right)}{25}}=11\)
\(\Leftrightarrow\sqrt[]{x+1}-2\sqrt[]{x+1}+\dfrac{16}{5}\sqrt[]{x+1}=11\)
\(\Leftrightarrow\dfrac{11}{5}\sqrt[]{x+1}=11\)
\(\Leftrightarrow\sqrt[]{x+1}=5\)
\(\Leftrightarrow x+1=25\)
\(\Leftrightarrow x=24\)
HQ
Hà Quang Minh
Giáo viên
17 tháng 9 2023
Nhớ viết thêm điều kiện vào nữa, ở đây điều kiện là \(x\ge-1\)
13 tháng 5 2020
\(\frac{x+9}{11}+\frac{x+23}{25}=\frac{x+6}{4}\)
\(\frac{100x+900}{1100}+\frac{44x+1012}{1100}=\frac{275x+650}{1100}\)
\(100x+900+44x+1012=275x+650\)
\(144x+1912=275x+650\)
\(144x+1912-275x-650=0\)
\(-131x+1262=0\)
\(-131x=-1262\)
\(x=\frac{1262}{131}\)
PN
20 tháng 5 2020
\(\frac{x+9}{11}+\frac{x+23}{25}=\frac{x+6}{4}\)
\(< =>\frac{\left(x+9\right).25+\left(x+23\right).11}{11.25}=\frac{x+6}{4}\)
\(< =>\frac{25x+11x+478}{275}=\frac{x+6}{4}\)
\(< =>\left(36x+478\right).4=\left(x+6\right).275\)
\(< =>144x+1912=275x+1650\)
\(< =>1912-1650=275x-144x=131x\)
\(< =>262=131x\)\(< =>x=\frac{262}{131}=2\)
2 tháng 2
a: 11x+4=-3/2
=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-9+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3+2\right)=0\)
=>(x-3)(x+5)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(3x-9+10x+5=90\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>2x=6
=>x=3(nhận)
PK
2
MH
16 tháng 1 2022
\(\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)
⇒ \(3x=-5\)
⇒ \(x=-\dfrac{5}{3}\)
16 tháng 1 2022
\(a,\left\{{}\begin{matrix}2x-2y=-4\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y+x+2y=\left(-4\right)+\left(-1\right)\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x=-5\\x+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\-\dfrac{5}{3}+2y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\2y=\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(b,\left\{{}\begin{matrix}3x+5y=11\\2x+5y=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=11\\3x+5y-2x-5y=11-9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3.2+5y=11\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}6+5y=11\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5y=5\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=2\end{matrix}\right.\)
31 tháng 5 2018
\(\dfrac{8}{x}-8+\dfrac{11}{x}-11=\dfrac{9}{x}-9+\dfrac{10}{x}-10\)\(\Leftrightarrow\dfrac{8}{x}+\dfrac{11}{x}-\dfrac{9}{x}-\dfrac{10}{x}=8+11-9-10\)
\(\Leftrightarrow\dfrac{8+11-9-10}{x}=0\)
\(\Leftrightarrow\dfrac{0}{x}=0\)
\(\Leftrightarrow x=0\)
S=\(\left\{0\right\}\)
18 tháng 5 2021
3(2x+y)-2(3x-2y)=3.19-11.2
6x+3y-6x+4y=57-22
7y=35
y=5
thay vào :
2x+y=19
2x+5=19
2x=14
x=7
2/ x2+21x-1x-21=0
x(x+21)-1(x+21)=0
(x+21)(x-1)=0
TH1 x+21=0
x=-21
TH2 x-1=0
x=1
vậy x = {-21} ; {1}
3/ x4-16x2-4x2+64=0
x2(x2-16)-4(x2-16)=0
(x2-16)-(x2-4)=0
TH1 x2-16=0
x2=16
<=>x=4;-4
TH2 x2-4=0
x2=4
x=2;-2
18 tháng 5 2021
Bài 1 :
\(\hept{\begin{cases}2x+y=19\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}4x+2y=38\\3x-2y=11\end{cases}\Leftrightarrow\hept{\begin{cases}7x=49\\2x+y=19\end{cases}}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=7\\2x+y=19\end{cases}}\)Thay vào x = 7 vào pt 2 ta được :
\(14+y=19\Leftrightarrow y=5\)Vậy hệ pt có một nghiệm ( x ; y ) = ( 7 ; 5 )
Bài 2 :
\(x^2+20x-21=0\)
\(\Delta=400-4\left(-21\right)=400+84=484\)
\(x_1=\frac{-20-22}{2}=-24;x_2=\frac{-20+22}{2}=1\)
Bài 3 : Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2-20t+64=0\)
\(\Delta=400+4.64=656\)
\(t_1=\frac{20+4\sqrt{41}}{2}\left(tm\right);t_2=\frac{20-4\sqrt{41}}{2}\left(ktm\right)\)
Theo cách đặt : \(x^2=\frac{20+4\sqrt{41}}{2}\Rightarrow x=\sqrt{\frac{20+4\sqrt{41}}{2}}=\frac{\sqrt{20\sqrt{2}+4\sqrt{82}}}{2}\)
13 tháng 10 2021
\(ĐK:-3\le x\le\dfrac{3}{2}\\ PT\Leftrightarrow11-x-4\sqrt{x+3}-2\sqrt{3-2x}=0\\ \Leftrightarrow\left(x+3-4\sqrt{x+3}+4\right)+\left(3-2x-2\sqrt{3-2x}+1\right)=0\\ \Leftrightarrow\left(\sqrt{x+3}-2\right)^2+\left(\sqrt{3-2x}-1\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{3-2x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=2\\3-2x=1\end{matrix}\right.\Leftrightarrow x=1\left(tm\right)\)
CB
0
Y
17 tháng 3 2023
\(\dfrac{3x}{2}+\dfrac{11}{x}=\dfrac{4}{3}\)\(\left(dkxd:x\ne0\right)\)
\(\Leftrightarrow\dfrac{3x}{2}+\dfrac{11}{x}-\dfrac{4}{3}=0\)
\(\Leftrightarrow\dfrac{3x.3x+11.6-4.2x}{6x}=0\)
\(\Leftrightarrow9x^2+66-8x=0\)
\(\Leftrightarrow9x^2-8x+66=0\)
\(\Leftrightarrow\) Pt vô nghiệm
\(\dfrac{11}{3}+\left|x\right|=\dfrac{9}{4}\)
=>\(\left|x\right|=\dfrac{9}{4}-\dfrac{11}{3}=\dfrac{27}{12}-\dfrac{44}{12}=-\dfrac{17}{12}\)
mà \(\left|x\right|>=0\forall x\)
nên \(x\in\varnothing\)