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a)(x − 12)2 = 0
=>x − 12 = 0
=> x = 12
b) (x+12)2 = 0,25
=> x + 12 = 0,5 hoặc x + 12= -0,5
=> x = -11,5 hoặc x = -12,5
c) (2x−3)3 = -8
=> 2x - 3 = -2
=> x = 0,5
d) (3x−2)5 = −243
=> 3x - 2 = -3
=> x = -1/3
e) (7x+2)-1 = 3-2
=> \(\dfrac{1}{7x+2}=\dfrac{1}{9}\)
=> 7x + 2 = 9
=> x = 1
f) (x−1)3 = −125
=> (x−1) = −5
=> x = -4
g) (2x−1)4 = 81
=> 2x - 1 = 3
=> x = 2
h) (2x−1)6 = (2x−1)8
=> 2x -1 = 0 hoặc 2x - 1 = 1 hoặc 2x - 1 = -1
=> x = 1/2 hoặc x = 1 hoặc x = 0
a/ \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
b/ \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{2}\right)^2\\\left(x+\dfrac{1}{2}\right)^2=\left(-\dfrac{1}{2}\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{2}\\x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy ..
c/ \(\left(2x-3\right)^3=-8\)
\(\Leftrightarrow\left(2x-3\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-3=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy ...
d/ \(\left(3x-2\right)^5=-243\)
\(\left(3x-2\right)^5=\left(-3\right)^5\)
\(\Leftrightarrow3x-2=-3\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy ...
e/ \(\left(x-1\right)^3=-125\)
\(\Leftrightarrow\left(x-1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x-1=-5\)
\(\Leftrightarrow x=-4\)
Vậy..
f/ \(\left(2x-1\right)^4=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=3^4\\\left(2x-1\right)^4=\left(-3\right)^4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy...
g/ \(\left(2x-1\right)^6=\left(2x-1\right)^8\)
\(\Leftrightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)
\(\Leftrightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\\left(2x-1\right)^2-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy..
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)
1) |x|=x+2
=> \(\left[{}\begin{matrix}x=x+2\\x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(voli\right)\\2x=-2\Rightarrow x=-1\end{matrix}\right.\)
vậy x=-1
c;b tương tự
2) \(\left|x-\dfrac{3}{2}\right|=\left|\dfrac{5}{2}-x\right|\)
=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{5}{2}-x\\x-\dfrac{3}{2}=x-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\Rightarrow x=2\\0=-1\left(voli\right)\end{matrix}\right.\)
vậy x=2
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x+2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=19\)
Chúc bạn học tốt!!!
a, \(\dfrac{x+1}{5}+\dfrac{x+3}{4}=\dfrac{x+5}{3}+\dfrac{x+7}{2}\)
\(\Rightarrow\dfrac{x+1}{5}+2+\dfrac{x+3}{4}+2=\dfrac{x+5}{3}+2+\dfrac{x+7}{2}+2\)
\(\Rightarrow\dfrac{x+11}{5}+\dfrac{x+11}{4}-\dfrac{x+11}{3}-\dfrac{x+11}{2}=0\)
\(\Rightarrow\left(x+11\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Vậy x = -11
b, \(\dfrac{3}{\left(x+2\right)\left(x+5\right)}+\dfrac{5}{\left(x+5\right)\left(x+10\right)}+\dfrac{7}{\left(x+10\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\dfrac{15}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow x=15\)
Vậy x = 15
a) \(\left|2,5-x\right|-1,3=0\)
th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)
th2: \(2,5-x< 0\Leftrightarrow x>2,5\)
\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)
vậy \(x=1,2;x=3,8\)
b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)
c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)
th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)
th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)
\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)
vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)
d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)
th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)
\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)
\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)
vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)
e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)
\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm
a. \(\dfrac{1}{3}.\left(x-1\right)+\dfrac{2}{5}.\left(x+1\right)=0\)
=> \(\dfrac{1}{3}x-\dfrac{1}{3}+\dfrac{2}{5}x+\dfrac{2}{5}=0\)
=> \(\dfrac{1}{3}x+\dfrac{2}{5}x=0+\dfrac{1}{3}-\dfrac{2}{5}\)
=> \(\dfrac{11}{15}x=\dfrac{-1}{15}\)
=> \(x=\dfrac{-1}{11}\)
Đây toán 8 mà? :v
a,\(\dfrac{1}{5}x\left(x-1\right)+\dfrac{2}{5}x\left(x+1\right)=0\)
\(\Leftrightarrow5x\left(x-1\right)+6x\left(x+1\right)=0\)
\(\Leftrightarrow\left[5\left(x-1\right)+6x\left(x+1\right)\right]x=0\)
\(\Leftrightarrow\left(5x-5+6x+6\right)x=0\)
\(\Leftrightarrow\left(11+1\right)x=0\)
\(\Leftrightarrow11x+1=0;x=0\)
\(\Leftrightarrow x=-\dfrac{1}{11};x=0\)
Vậy....
Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
\(a,\left(2-x\right)\left(\dfrac{4}{5}-x\right)< 0\)
=>Trong 2 số phải có 1 số âm và 1 số dương
Mà \(2-x>\dfrac{4}{5}-x\)
=>\(\dfrac{4}{5}< x< 2\)
Vậy...
a: =>|3/2x|=-2+0,4+0,6=-1(vô lý)
b: =>|x+7/3|=1/3
=>x+7/3=1/3 hoặc x+7/3=-1/3
=>x=-2 hoặc x=-8/3