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ap dung cong thuc: a/b = c/d <=> ad= bc <=> c = ad/b
A = (4x2-7x+3)(x2+2x+1)/(x2-1)
\(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(x^3-3^3+x\left(2^2-x^2\right)=1\)
\(x^3-27+4x-x^3=1\)
\(4x-27=1\)
\(4x=28\)
\(x=7\)
Vậy x = 7
a) \(\left(7-14x\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}7-14x=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}14x=7-0\\x=2\end{cases}\Leftrightarrow}}\orbr{\begin{cases}14x=7\\x=2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=2\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{2};2\right\}\)
b) \(\left(2x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-1\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}}\)
Vậy \(x\in\left\{-\frac{1}{2};3\right\}\)
+) (5x-1). (2x+3)-3. (3x-1)=0
10x^2+15x-2x-3 - 9x+3=0
10x^2 +8x=0
2x(5x+4)=0
=> x=0 hoặc x= -4/5
+) x^3 (2x-3)-x^2 (4x^2-6x+2)=0
2x^4 -3x^3 -4x^4 + 6x^3 - 2x^2=0
-2x^4 + 3x^3-2x^2=0
x^2(-2x^2+x-2)=0
-2x^2(x-1)^2=0
=> x=0 hoặc x=1
+) x (x-1)-x^2+2x=5
x^2 -x -x^2+2x=5
x=5
+) 8 (x-2)-2 (3x-4)=25
8x - 16-6x+8=25
2x=33
x=33/2
\(P=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
Vì \(\left(x-1\right)^2\ge0\Rightarrow\left(x-1\right)^2+4\ge4\)
=>Pmin=(x-1)2+4=4
<=>(x-1)2=0
<=>x-1=0
<=>x=1
Vậy Pmin=4 khi x=1
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\(Q=2x^2-6x=2\left(x^2-3x\right)=2\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]-\frac{9}{2}=2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}\ge-\frac{9}{2}\)
=>Qmin=\(2\left(x-\frac{3}{2}\right)^2-\frac{9}{2}=-\frac{9}{2}\)
<=>\(2\left(x-\frac{3}{2}\right)^2=0\)
<=>\(\left(x-\frac{3}{2}\right)^2=0\)
<=>\(x-\frac{3}{2}=0\)
<=>\(x=\frac{3}{2}\)
Vậy Qmin=\(-\frac{9}{2}\) khi \(x=\frac{3}{2}\)
-2x + x = (-2 + 1)x = -x em nhé
Không phải 2x + x = x nhé