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1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
a)\(\frac{x+3}{x+5}=7\Leftrightarrow x+3=7\left(x+5\right)\)
\(\Leftrightarrow x+3=7x+35\)
\(\Leftrightarrow-6x=32\)
\(\Leftrightarrow x=-\frac{16}{3}\)
b)\(\frac{2x-1}{3x+5}=-\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-1\right)=-2\left(3x+5\right)\)
\(\Leftrightarrow6x-3=-6x-10\)
\(\Leftrightarrow12x=-7\)
\(\Leftrightarrow x=-\frac{7}{12}\)
c)\(\frac{x+1}{4}=\frac{9}{x+1}\Leftrightarrow\left(x+1\right)^2=36\)
\(\Leftrightarrow\left(x+1\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}}\)
d)\(\frac{6x-1}{2x+3}=\frac{3x}{x+2}\)
\(\Leftrightarrow\left(6x-1\right)\left(x+2\right)=3x\left(2x+3\right)\)
\(\Leftrightarrow6x^2+12x-x-2=6x^2+9x\)
\(\Leftrightarrow2x=2\Leftrightarrow x=1\)
A=5-3(2x+1)^2
Ta có : (2x+1)^2\(\ge\)0
\(\Rightarrow\)-3(2x-1)^2\(\le\)0
\(\Rightarrow\)5+(-3(2x-1)^2)\(\le\)5
Dấu = xảy ra khi : (2x-1)^2=0
=> 2x-1=0 =>x=\(\frac{1}{2}\)
Vậy : A=5 tại x=\(\frac{1}{2}\)
Ta có : (x-1)^2 \(\ge\)0
=> 2(x-1)^2\(\ge\)0
=>2(x-1)^2+3 \(\ge\)3
=>\(\frac{1}{2\left(x-1\right)^2+3}\)\(\le\)\(\frac{1}{3}\)
Dấu = xảy ra khi : (x-1)^2 =0
=> x = 1
Vậy : B = \(\frac{1}{3}\)khi x = 1
\(\frac{x^2+8}{x^2+2}\)= \(\frac{x^2+2+6}{x^2+2}=1+\frac{6}{x^2+2}\)
Làm như câu B GTNN = 4 khi x =0
k vs nha
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
a)\(\frac{1}{4}-\frac{1}{3}x=\frac{2}{5}-\frac{3}{2}x\)
\(\Leftrightarrow\)\(\frac{15-20x}{60}=\frac{24-90x}{60}\)
\(\Leftrightarrow15-20x=24-90x\)
\(\Leftrightarrow-20x+90x=24-15\)
\(\Leftrightarrow70x=9\)
\(\Leftrightarrow x=\frac{9}{70}\)
c) (1/2-1/6)*3^x+4-4*3^x=3^16-4*3^13
=1/3*3^x*3^4-4*3^x=3^13*3^3-4*3^13
=27*3^x-4*3^x=3^13*(27-4)
=3^x*(27-4)=3^13*(27-4)
=>x=13
Bài 1 :
a. \(\left|x-\frac{1}{3}\right|< \frac{5}{2}\)
TH1 : nếu \(\left|x-\frac{1}{3}\right|>0\)
\(x-\frac{1}{3}< \frac{5}{3}\)
\(x< 2\)
TH2 : nếu \(\left|x-\frac{1}{3}\right|< 0\)
\(\frac{1}{3}-x< \frac{5}{3}\)
\(x>-\frac{4}{3}\)
Bài 2 :
a. \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2-1=0\)
\(\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\left(x-3\right)\left(x-1\right)=0\)
\(\left[\begin{array}{nghiempt}x-3=0\\x-1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
Giúp mình bài này với ạ
Tìm x
a) 3(x-4)+2,6=5,9
b)|1+x|+3=3
c)(\(\frac{-2}{3}\))\(^x\)=\(\frac{-8}{27}\)
a) 3(x-4)+2,6=5,9
\(3\left(x-4\right)=3,3\)
\(x-4=\frac{11}{10}\)
\(x=\frac{51}{10}\)
b)|1+x|+3=3
\(\Rightarrow|1+x|=0\)
\(1+x=0\)
\(x=-1\)
c) \(\left(\frac{-2}{3}\right)^x=\frac{-8}{27}\)
\(\left(\frac{-2}{3}\right)^x=\left(\frac{-2}{3}\right)^3\)
\(x=3\)
chúc bạn học tốt
a) 3x+3x+2=812
Suy ra 3x+3x.32=812
3x.(1+32) =812
3x.10 =812
3x =812:10
3x =406/5
Suy ra x ko có giá trị
b)4\(\frac{1}{3}\):\(\frac{x}{4}\)=6:0,3
suy ra \(\frac{13}{3}\):\(\frac{x}{4}\) =20
x/4 = 13/3:20
x/4 = 13/60
x = 13/15
c) I 2x + 0,5I=8,5
2x+0,5=8,5 hoặc 2x+0,5=-8,5
TH1:2x+0,5=8,5=>x=4
TH2:2x+0,5=-8,5=>x=-9/2
d) 8x: 2x =1635
=>(8:2)x=1635
=>4x =1635
=>4x =(42)35
=>4x =42.35 =>4x=470 =>x=70
Vậy x = 70
Đầy đủ và chính xác lắm đó.
Bạn Thanh thấy đáp án chưa?