l \(2x-1\)l = \(\left(-4\right)^2\)
l l: giá trị tuyệt đối
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\(\left|2-x\right|+\left|x+1\right|=5\)
TH1 : \(\left|2-x\right|=\pm5\)
+ ) \(2-x=5\)
\(x=2-5\)
\(x=-3\)
+ ) \(2-x=\left(-5\right)\)
\(x=2-\left(-5\right)\)
\(x=7\)
TH2 : \(\left|x+1\right|=\pm5\)
+ ) \(x+1=5\)
\(x=5-1\)
\(x=4\)
+ ) \(x+1=\left(-5\right)\)
\(x=\left(-5\right)-1\)
\(x=-6\)
2 ) \(\left|x+1\right|+\left|2x+1\right|=22\)
TH1 : \(\left|x+1\right|=\pm22\)
+ ) \(x+1=22\)
\(x=22-1\)
\(x=21\)
+ ) \(x+1=-22\)
\(x=-22-1\)
\(x=-23\)
TH2: \(\left|2x+1\right|=\pm22\)
+ ) \(2x+1=22\)
\(2x=21\)
\(x=\frac{21}{2}\)
+ ) \(2x+1=-22\)
\(2x=-23\)
\(x=\frac{-23}{2}\)
a) \(x>3\Leftrightarrow\left|x-3\right|=x-3\)
\(x< 3\Leftrightarrow\left|x-3\right|=3-x\)
b) \(x>0\Leftrightarrow\left|-2x\right|=2x\)
\(x< 0\Leftrightarrow\left|-2x\right|=-2x\)
ta rút gọn và nhân thì ta được (4\2-7\12-9\15):(4\3-1\2-5\3)
=(120\60-35\60-36\60):(8\6-3\6-10\6)
=49\60:-5\6
= 49\-300
chúc bạn học tốt !
(+) Nếu -2x+4>=0 <=> -2x >= -4<=> x<= -2 thì |-2x+4| = -2x+4:
Ta có pt: -2x+4-2(x+1)=-5x+1 <=> -2x+4-2x-2+5x-1=0 <=> x+1=0 <=> x=-1 (Ko thỏa mãn đk)
(+) Nếu -2x+4<0 <=> -2x<-4 <=> x>-2 thì |-2x+4|=-(-2x+4)=2x-4:
Ta có pt: 2x-4-2(x+1)=-5x+1 <=> 2x-4-2x-2+5x-1 =0 <=> 5x-7=0 <=> x= 7/5 (Thỏa mãn đk)
Vay tap nghiem cua pt la S={7/5}
\(\left|x+\frac{1}{2}\right|-2x=3\)
<=>\(\left|x+\frac{1}{2}\right|=3+2x\)
<=>\(x+\frac{1}{2}=-\left(3+2x\right)\)hoặc\(3+2x\)
Xét \(x+\frac{1}{2}=-\left(3+2x\right)\)
<=>\(x+\frac{1}{2}=3-2x\)
<=>\(x=\frac{5}{6}\left(Loai\right)\)
Xét \(x+\frac{1}{2}=3+2x\)
<=>\(x=-\frac{7}{6}\left(tm\right)\)
Vậy \(x=-\frac{7}{6}\)
\(\left|x-\frac{1}{2}\right|-2x=3\)
\(\Rightarrow\left[\begin{array}{nghiempt}x-\frac{1}{2}-2x==3\\\frac{1}{2}-x-2x=3\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}-x=\frac{7}{2}\\-3x=\frac{5}{2}\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=-\frac{7}{2}\\x=-\frac{5}{6}\end{array}\right.\)
\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\)
vì \(\left|x+y-z\right|=95\Rightarrow\orbr{\begin{cases}x+y-z=95\\x+y-z=-95\end{cases}}\)
th1: x+y-z=95
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=150\)
\(\frac{x}{\frac{1}{2}}=150\Rightarrow x=75\)
\(\frac{y}{\frac{1}{3}}=150\Rightarrow y=50\)
\(\frac{z}{\frac{1}{5}}=150\Rightarrow z=30\)
th2: x+y-z=-95
\(\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=-\frac{95}{\frac{19}{30}}=-150\)
\(\frac{x}{\frac{1}{2}}=-150\Rightarrow x=-75\)
\(\frac{y}{\frac{1}{3}}=-150\Rightarrow y=-50\)
\(\frac{z}{\frac{1}{5}}=-150\Rightarrow z=-30\)
vậy x=75, y=50,z=30
hay x=-75, y=-50, x=-30
\(\left|2x-1\right|=\left(-4\right)^2\)
\(\left|2x-1\right|=16\)
\(\Rightarrow\orbr{\begin{cases}2x-1=16\\2x-1=-16\end{cases}\Rightarrow\orbr{\begin{cases}2x=17\\2x=-15\end{cases}\Rightarrow}\orbr{\begin{cases}x=8,5\\x=-7,5\end{cases}}}\)
vậy....................
=> \(\orbr{\begin{cases}2x-1=-4\\2x-1=4\end{cases}}\)
=> \(\orbr{\begin{cases}x=-3,5\\x=2,5\end{cases}}\)