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Bài 46:
11: Ta có: \(-4\left|x-2\right|=-8\)
\(\Leftrightarrow\left|x-2\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy: x∈{0;4}
12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)
\(\Leftrightarrow5\left|x+2\right|=20\)
\(\Leftrightarrow\left|x+2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)
Vậy: x∈{-6;2}
13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)
\(\Leftrightarrow6\left|x-2\right|=-6\)(1)
Ta có: \(\left|x-2\right|\ge0\forall x\)
\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)
Ta có: -6<0(3)
Từ (1), (2) và (3) suy ra x∈∅
Vậy: x∈∅
14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)
\(\Leftrightarrow-7\left|x+4\right|=-7\)
\(\Leftrightarrow\left|x+4\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy: x∈{-5;-3}
15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)
\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)
\(\Leftrightarrow4\left|x+1\right|=24\)
\(\Leftrightarrow\left|x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{-7;5}
16: Ta có: \(3\left|x+5\right|=-9\)(4)
Ta có: |x+5|≥0∀x
⇒3|x+5|≥0∀x(5)
Ta có: -9<0(6)
Từ (4), (5) và (6) suy ra x∈∅
Vậy: x∈∅
17: Ta có: \(-8\left|x-3\right|=24-16:2\)
\(\Leftrightarrow-8\left|x-3\right|=16\)
\(\Leftrightarrow\left|x-3\right|=-2\)
mà |x-3|≥0>-2∀x
nên x∈∅
Vậy: x∈∅
18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)
\(\Leftrightarrow-3\left|x+6\right|=3\)
\(\Leftrightarrow\left|x+6\right|=-1\)
mà |x+6|≥0>-1∀x
nên x∈∅
Vậy: x∈∅
19: Ta có: \(5-\left|x+7\right|=4\)
\(\Leftrightarrow\left|x+7\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)
Vậy: x∈{-8;-6}
20: Ta có: \(12-\left|x+8\right|=10\)
\(\Leftrightarrow\left|x+8\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)
Vậy: x∈{-10;-6}
-5/8 = x / 16
x = 16 : -5/8
x= -25,6
Y/10 = (-4)/8
Y = (-4)/8 x 10
Y = -5
\(\frac{-5}{8}=\frac{x}{16}\)
\(=>8x=-5.16\)
\(=>8x=-80=>x=-10\)
\(\frac{y}{10}=\frac{-4}{8}\)
\(=>\frac{y}{10}=\frac{-1}{2}\)
\(=>2y=-10\)
\(=>x=-5\)
\(\frac{x}{-4}=\frac{-16}{x}\)
\(\Rightarrow x\cdot x=-16\cdot\left(-4\right)\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x^2=\left(\pm8\right)^2\)
\(\Rightarrow x=\pm8\)
a)\(\left(x-\frac{5}{8}\right).\frac{5}{18}=-\frac{15}{36}\)
\(\Rightarrow x-\frac{5}{8}=\frac{-15}{36}.\frac{18}{5}\)
\(\Rightarrow x-\frac{5}{8}=-\frac{3}{2}\)
\(\Rightarrow x=-\frac{12}{8}+\frac{5}{8}=-\frac{7}{8}\)
b)\(\frac{x}{-4}=\frac{-16}{x}\)
\(\Rightarrow x^2=64\)
\(\Rightarrow x=\orbr{\begin{cases}8\\-8\end{cases}}\)
58 x 32 + 32 x 8 + 16 x 5
= 32 x ( 58 + 32 ) + 16 x 5
=32 x 90 + 16 x 5
= 16 x 2 x 90 + 16 x 5
= 16 x 180 + 16 x 5
= 16 x ( 180 + 5 )
= 16 x 185
= 2960
\(58.32+32.8+16.5\)
\(=16.116+16.16+16.5\)
\(=16.\left(116+16+5\right)\)
\(=16.137\)
\(=2192\)
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
\(\text{ a, 5-(10x)=-7}\)
\(\Rightarrow\) 10x=5-(-7)
\(\Rightarrow\) 10x=12
\(\Rightarrow\) x=12:10
\(\Rightarrow\) x=1,2
b, (x-5).(2x+8)=0
\(\Rightarrow\) x-5=0 hoặc 2x+8=0
\(\Rightarrow\) x =0+5 \(\Rightarrow\) 2x =0+8
\(\Rightarrow\) x =5 \(\Rightarrow\) 2x =8
\(\Rightarrow\) x =8:2
\(\Rightarrow\) x =4
vậy x\(\in\){5;4}
c, 2x-9=-8+9
\(\Rightarrow\) 2x-9=1
\(\Rightarrow\) 2x =1+9
\(\Rightarrow\) 2x =10
\(\Rightarrow\) x =10:2
\(\Rightarrow\) x =5
d, |x-9|.(-8)=-16
\(\Rightarrow\)|x-9| =-16:(-8)
\(\Rightarrow\)|x-9| =2
\(\Rightarrow\) x-9 =\(\hept{\begin{cases}2\\-2\end{cases}}\)
trường hợp 1: x-9=2
\(\Rightarrow\) x =2+9
\(\Rightarrow\) x =11
trường hợp 2: x-9=-2
\(\Rightarrow\) x =-2+9
\(\Rightarrow\) x =7
vậy x \(\in\){11;7}
# học tốt #
tim x