x^2-5x+12 bằng bn ạ
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T
8 tháng 10 2023
\(a)24\times(x-16)=12^2\\\Rightarrow 24\times(x-16)=144\\\Rightarrow x-16=144:24\\\Rightarrow x-16=6\\\Rightarrow x=6+16\\\Rightarrow x=22\\---\\b)(x^2-10):5=5\\\Rightarrow x^2-10=5\times5\\\Rightarrow x^2-10=25\\\Rightarrow x^2=25+10\\\Rightarrow x^2=35\\\Rightarrow x=\pm\sqrt{35}\\---\)
\(c)(5x+335):2=400\\\Rightarrow 5x+335=400\times2\\\Rightarrow 5x+335=800\\\Rightarrow 5x=800-335\\\Rightarrow 5x=465\\\Rightarrow x=465:5\\\Rightarrow x=93\\---\\d)63:(5x+4)=2^3-1\\\Rightarrow 63:(5x+4)=8-1\\\Rightarrow 63:(5x+4)=7\\\Rightarrow 5x+4=63:7\\\Rightarrow 5x+4=9\\\Rightarrow 5x=9-4\\\Rightarrow 5x=5\\\Rightarrow x=5:5\)
\(\Rightarrow x=1\)
\(Toru\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 6 2023
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
8 tháng 9 2023
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
LB
9 tháng 4 2020
Ta có:
\(\left(2x-1\right)^2+\left(x-2\right)\left(x+2\right)=\left(5x+1\right)\left(x-4\right)-12\)
\(\left(4x^2-4x+1\right)+x^2-4=5x^2-19x-4-12\)
\(5x^2-4x-3=5x^2-19x-16\)
\(\left(5x^2-5x^2\right)+\left(19x-4x\right)+\left(16-3\right)=0\)\(15x+13=0\)\(x=-\frac{13}{15}\)
JS
1
10 tháng 4 2020
Toán lớp 8 nha mn mình ấn lộn
Mik cần gấp nha mn💛🙆
ND
1
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 2 2020
\(\Leftrightarrow\left(x^4+x^3-2x^2\right)+\left(x^3+x^2-2x\right)+\left(6x^2+6x-12\right)=0\)
\(\Leftrightarrow x^2\left(x^2+x-2\right)+x\left(x^2+x-2\right)+6\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=0\left(vn\right)\\x^2+x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(=x^2-2.\frac{5}{2}x+\frac{25}{4}+\frac{23}{4}\)
\(=\left(x-\frac{5}{2}\right)^2+\frac{23}{4}\)