Tìm x
4.(2x+7)-3.(3x-2)=24
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\(1,3x-7=19\\ \Rightarrow3x=26\\ \Rightarrow x=\dfrac{26}{3}\\ 2,\left(2x+1\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x+1=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\\ 3,3x+\dfrac{2}{4}+1=5x-\dfrac{1}{3}\\ \Rightarrow5x-\dfrac{1}{3}-3x-\dfrac{2}{4}-1=0\\ \Rightarrow2x-\dfrac{11}{6}=0\\ \Rightarrow2x=\dfrac{11}{6}\\ \Rightarrow x=\dfrac{11}{12}\)
\(4,\dfrac{x}{15}+\dfrac{1}{2}-\dfrac{x}{50}=\dfrac{5}{6}\\ \Rightarrow\dfrac{x}{15}-\dfrac{x}{50}=\dfrac{5}{6}-\dfrac{1}{2}\\ \Rightarrow x\left(\dfrac{1}{15}-\dfrac{1}{50}\right)=\dfrac{1}{3}\\ \Rightarrow\dfrac{7}{150}x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{50}{7}\)
\(a,\left(3x-7\right)^2=\left(2-2x\right)^2\)
a,\(=>\left(3x-7\right)^2-\left(2-2x\right)^2=0\)
\(< =>\left(3x-7+2-2x\right)\left(3x-7-2+2x\right)=0\)
\(< =>\left(x-5\right)\left(5x-9\right)=0=>\left[{}\begin{matrix}x=5\\x=1,8\end{matrix}\right.\)
b, \(x^2-8x+6=0< =>x^2-2.4x+16-10=0\)
\(< =>\left(x-4\right)^2-\sqrt{10}^2=0\)
\(=>\left(x-4+\sqrt{10}\right)\left(x-4-\sqrt{10}\right)=0\)
\(=>\left[{}\begin{matrix}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{matrix}\right.\)
c, \(4x^2-2x-1=0\)
\(< =>\left(2x\right)^2-2.2.\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(=>\left(2x-\dfrac{1}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)
\(=>\left(2x+\dfrac{-1+\sqrt{5}}{2}\right)\left(2x-\dfrac{1+\sqrt{5}}{2}\right)=0\)
\(=>\left[{}\begin{matrix}x=\dfrac{1-\sqrt{5}}{4}\\x=\dfrac{1+\sqrt{5}}{4}\end{matrix}\right.\)
d,\(x^4-4x^2-32=0\)
đặt \(t=x^2\left(t\ge0\right)=>t^2-4t-32=0\)
\(< =>t^2-2.2t+4-6^2=0\)
\(=>\left(t-2\right)^2-6^2=0=>\left(t-8\right)\left(t+4\right)=0\)
\(=>\left[{}\begin{matrix}t=8\left(tm\right)\\t=-4\left(loai\right)\end{matrix}\right.\)\(=>x=\pm\sqrt{8}\)
4(2x+7)-3(3x-2)=24
4.2x+4.7-3.3x+3.2=24
8x+28-9x+6=24
8x-9x=24-28-6=-10
=>(-1)x=-10
x=-10:(-1)
x=10
cứ tin mình,mình học loại này rồi
a)
4 . ( 2x + 7 ) - 3 . ( 3x - 2 ) = 24
8x + 28 - 9x + 6 = 24
34 - x = 24
x = 10
b)
3 ( x - 2 ) + 2x = 10
3x - 6 + 2x = 10
5x - 6 = 10
5x = 16
x = 16/5
a) 4 . ( 2x + 7 ) - 3 . ( 3x - 2 ) = 24
\(8x+28-9x+6=24\)
\(8x-9x=24-6-28\)
\(-x=-10\)
\(x=10\)
b) 3 ( x - 2 ) + 2x = 10
\(3x-6+2x=10\)
\(3x+2x=10+6\)
\(5x=16\)
\(x=\frac{16}{5}\)
học tốt !!!
4(2x + 7) - 3(3x - 2) = 24
=> 8x + 28 - 9x + 6 = 24
=> -x + 34 = 24
=> -x = 24 - 34
=> -x = -10
=> x = 10
Học tốt
a,6x-3-5x+15+18x-24=24
19x-12=24
19x=36
x=36/19
c,10x-6x2+6x2-10x+21=3
0x=-18
không có x
d,3x2+3x-2x2-4x=-1-x
x2-x=-1-x
x2-x+x=-1
x2=-1
không có x thỏa mãn
4.2x+4.7-3.3x+3.2=24
8x+28-9x+6=24
(8x-9x)+(28+6)=24
-1x+34=24
-1x=24-34
-1x=10
x=10:(-1)
x=-10
Vậy x = -10
\(4\left(2x+7\right)-3\left(3x-2\right)=24\)
\(8x+28-9x+6=24\)
\(-x+34=24\)
\(-x=-10\)
\(x=10\)
4.(2x+7)-3.(3x-2)=24
4.2x+4.7-3.3x-3.2=24
8x+28-9x+6=24
8x-9x=24-28-6=-10
\(\Rightarrow\)(-1)x=-10
x=-10:(-1)
x=10
Vậy x=10