1.tìm x biết :
a.3x + 5 = 8
b.1/2x+ 9/4 = 17/4
c. (x-1)² = 1/9
d.|1/2x - 1/3| = 1/6
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2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
tìm x biết:
(3x-1) [- 1/2x+5]=0
1/4+1/3:(2x-1)=-5
[2x+3/5]2 - 9/25=0
-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6
[x+1/2]x [2/3-2x]=0
17/2-|2x-3/4|=-7/4
2/3x-1/2x =5/12
(x+1/5)2+17/25=26/25
[x.44/7+3/7].11/5-3/7=-2
3[3x-1/2]+1/9=0
Toán lớp 6Tìm x
Trả lời Câu hỏi tương tự
Chưa có ai trả lời câu hỏi này,bạn hãy là người đâu tiên giúp nguyenvanhoang giải bài toán này !
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
a)\(2x\left(x+1\right)-3-2x=5\)
\(\Leftrightarrow2x^2+2x-3-2x=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4=\left(-2\right)^2=2^2\)
\(\Rightarrow x=2;-2\)
b)\(2x\left(3x+1\right)+\left(4-2x\right)=7\)
\(\Leftrightarrow6x^2+2x+4-2x=7\)
\(\Leftrightarrow6x^2+4=7\)
\(\Leftrightarrow6x^2=3\)
\(\Leftrightarrow x^2=\frac{1}{2}=-\sqrt{\frac{1}{2}}=\sqrt{\frac{1}{2}}\)
c)\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x-1\right)^2=6\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x^2-2x+1\right)=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow-3x^2+27x+6x^2-12x+6=6\)
\(\Leftrightarrow3x^2+15x=0\)
\(\Leftrightarrow3x\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x+5=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
a.
\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)
b.
\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)
c.
\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
d.
\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
e.
\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)
f.
\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)
1.
a. 3x+5=8
3x = 8-5
3x = 3
x = 3:3
x = 1
b.1/2x+9/4=17/4
1/2x = 17/4 - 9/4
1/2x = 2
x = 2:1/2
x = 4
c. ( x-1)2 = 1/9
(x-1)2 = (1/3)2
x-1 = 1/3
x =1/3 +1
x = 4/3
d . | 1/2x - 1/3 | = 1/6
suy ra 1/2x - 1/3 =1/6 hoặc 1/2x - 1/3 = -1/6
1/2x= 1/6 + 1/3 hoặc 1/2x = -1/6 +1/3
1/2x= 1/2 hoặc 1/2x = 1/6
x = 1/2 : 1/2 hoặc x = 1/6 : 1/2
x =1 hoặc x = 1/3
vậy x E \(\hept{\begin{cases}\\\end{cases}}1;\frac{1}{3}\)
a, 3x + 5 = 8
3x = 8 - 5
3x = 3
x = 3:3
x = 1
b, \(\frac{1}{2}\)x + \(\frac{9}{4}\)=\(\frac{17}{4}\)
\(\frac{1}{2}x=\frac{17}{4}-\frac{9}{4}\)
\(\frac{1}{2}x=2\)
\(x=2:\frac{1}{2}\)
x = 4
c, (x - 1)2 = \(\frac{1}{9}\)
=> (x - 1)2 = \(\left(\frac{1}{3}\right)^2\)hoặc (x - 1)2 = \(\left(-\frac{1}{3}\right)^2\)
x - 1 = \(\frac{1}{3}\)hoặc x - 1 = \(-\frac{1}{3}\)
x = \(\frac{1}{3}+1\)hoặc x = \(-\frac{1}{3}+1\)
x = \(\frac{4}{3}\)hoặc x = \(\frac{2}{3}\)
d, \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{6}\)
=> \(\frac{1}{2}x-\frac{1}{3}=\frac{1}{6}\)hoặc \(\frac{1}{2}x-\frac{1}{3}=-\frac{1}{6}\)
\(\frac{1}{2}x=\frac{1}{6}+\frac{1}{3}\)hoặc \(\frac{1}{2}x=-\frac{1}{6}+\frac{1}{3}\)
\(\frac{1}{2}x=\frac{1}{2}\)hoặc \(\frac{1}{2}x=\frac{1}{6}\)
\(x=\frac{1}{2}:\frac{1}{2}\)hoặc \(x=\frac{1}{6}:\frac{1}{2}\)
x = 1 hoặc x = \(\frac{1}{3}\)