giải hệ pt
x mũ 2 -7x +6=0
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\(\text{7x - x = 5}^{21}:5^{19}+3.2^2-7^0\)
\(\text{(7-1)x=5}^2\text{ + 3. 4 - 1}\)
\(\text{6x = 25 + 12 - 1}\)
\(\text{6x = 36}\)
\(\text{ x = 6}\)
a)7.x-x=521 :519 +3.22 -70
x.(7-1)=52+12-1
x.6 =36
x =36:6=6
b)7.x-2.x=617:615+44:11
x.(7-2)=62+4
x.5 =40
x=40:5=8
16. 7 x − 16. 5 2 x = 0
⇔ 7 x = 5 2 x ⇔ 7 / 25 x = 7 / 25 0 ⇔ x = 0
Bài 5 :
f, bạn xem lại đề hay là tìm x chứa tham số a ?
g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow x=-3;x=2\)
h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)
\(\Leftrightarrow\left(5-x\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=5\)
m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)
n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)
\(\left\{{}\begin{matrix}x+y=80\\2x+3y=198\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3y=240\\2x+3y=198\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=80\\x=240-198=42\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=42\\y=38\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x+y=80\\2x+3y=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}2x+2y=160\\2x+3y=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\2x+3\cdot38=198\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\2x=84\end{matrix}\right.\)
⇔\(\left\{{}\begin{matrix}y=38\\x=42\end{matrix}\right.\)
Vậy (42;38) là nghiệm
\(\left\{{}\begin{matrix}x-2y=3\\2x+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\2\left(3+2y\right)+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\6+4y+3y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2y\\7y=-7\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3+2\left(-1\right)\\y=-1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=6\\2x+3y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=3+2y=3-2=1\end{matrix}\right.\)
Đặt H \(=x^4-5x^3+7x^2-6\)
Gỉa sử : \(H=\left(x^2+ax+b\right)\left(x^2+cx+d\right)\)
\(=x^4+cx^3+dx^2+ax^{3\:}+acx^2+adx+bx^2+bcx+bd\)
\(=x^4+\left(a+c\right)x^3+\left(ac+b+d\right)x^2+\left(ad+bc\right)x+bd\)
\(\Leftrightarrow\hept{\begin{cases}a+c=-5\\ac+b+d=7\\ad+bc=0\end{cases}}\)
\(\left\{bd=6\right\}\)
\(\Leftrightarrow\hept{\begin{cases}a=-3\\b=3\\c=-2\end{cases}}\)
\(\left\{d=-2\right\}\)
\(\Rightarrow H=\left(x^2-3x+3\right)\left(x^2-2x-2\right)\)
Chúc bạn học tốt !!!
1. 2x=16\(\Rightarrow\)X=4
2. 22x-1=27
\(\Rightarrow\)27=22.4-1
Vậy x =4
Đặt \(x^3=y\)
Khi đó pt trở thành \(y^2-7y+6=0\)
\(\Leftrightarrow y^2-6y-y+6=0\)
\(\Leftrightarrow\left(y-6\right)\left(y-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}y-6=0\\y-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}y=6\\y=1\end{cases}}\)
\(\left(+\right)y=1\Rightarrow x^3=1\Leftrightarrow x=1\)
\(\left(+\right)y=6\Rightarrow x^3=6\Leftrightarrow x=\sqrt[3]{6}\)
Vậy phương trình có nghiệm \(x=1;x=\sqrt[3]{6}\)
x²-7x+6=0
x.x-7.x=-6
x(x-7)=-6
Vì x€ Z nên x-7€ Z
Mà x( x-7)=-6
=>-6 chia hết cho x
Hay x là ước của -6
=>x€{1,2,3,6,-1,-2,-3,-6}
Vậy x€{1,2,3,6,-1,-2,-3,-6}
\(x^2-7x+6=0\\ x^2-x-6x+6=0\\ \left(x-6\right)\left(x-1\right)=0\\ =>\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)