tim x
a/ 53x+6 . 1257 = 62539
b/ 7. 3x+2 + 4. 3x+1 = 32014.2 - 32011. 29
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a: Ta có: \(4x\left(x-7\right)-4x^2=56\)
\(\Leftrightarrow4x^2-7x-4x^2=56\)
hay x=-8
b: Ta có: \(12x\left(3x-2\right)-\left(4-6x\right)=0\)
\(\Leftrightarrow36x^2-24x-4+6x=0\)
\(\Leftrightarrow36x^2-18x-4=0\)
\(\text{Δ}=\left(-18\right)^2-4\cdot36\cdot\left(-4\right)=900\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{18-30}{72}=\dfrac{-1}{6}\\x_2=\dfrac{18+30}{72}=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(4\left(x-5\right)-\left(x-5\right)^2=0\)
\(\Leftrightarrow\left(x-5\right)\left(4-x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=9\end{matrix}\right.\)
a: Ta có: \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)
c: Ta có: \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-5\right)\cdot\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b.
PT $\Leftrightarrow (5x^2-2x+10)^2-(3x^2+10x-8)^2=0$
$\Leftrightarrow (5x^2-2x+10-3x^2-10x+8)(5x^2-2x+10+3x^2+10x-8)=0$
$\Leftrightarrow (2x^2-12x+18)(8x^2+8x+2)=0$
$\Leftrightarrow (x^2-6x+9)(4x^2+4x+1)=0$
$\Leftrightarrow (x-3)^2(2x+1)^2=0$
$\Leftrightarrow (x-3)(2x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $2x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-\frac{1}{2}$
d.
$x^2-2x=24$
$\Leftrightarrow x^2-2x-24=0$
$\Leftrightarrow (x+4)(x-6)=0$
$\Leftrightarrow x+4=0$ hoặc $x-6=0$
$\Leftrightarrow x=-4$ hoặc $x=6$
\(a,\dfrac{3}{4}x-\dfrac{2}{3}x=\dfrac{2}{7}\times\dfrac{1}{6}+\dfrac{5}{7}\times\dfrac{1}{6}\)
\(\dfrac{9}{12}x-\dfrac{8}{12}x=\dfrac{1}{6}\times\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(\dfrac{1}{12}x=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}:\dfrac{1}{12}\)
\(x=\dfrac{1}{6}\times12\)
\(x=2\)
a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow80x=480\Rightarrow x=6\)
b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow4x=0\Rightarrow x=0\)
c) \(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)
a) |2x - 5| + |3x + 1| = 6
Ta có: (5 - 2x) + (-3x - 1) = 6
=> 4 - 5x = 6
=> 5x = 4 - 6 = -2
\(\Rightarrow x=\frac{-2}{5}\), thỏa mãn \(x< \frac{-1}{3}\)
Ta có: (5 - 2x) + (3x + 1) = 6
=> 6 + x = 6
=> x = 6 - 6 = 0, thỏa mãn \(\frac{-1}{3}\le x< \frac{5}{2}\)
Ta có: (2x - 5) + (3x + 1) = 6
=> 5x - 4 = 6
=> 5x = 6 + 4 = 10
=> x = 10 : 5 = 2, không thỏa mãn \(x\ge\frac{5}{2}\)
Vậy \(\left[\begin{array}{nghiempt}x=\frac{-2}{5}\\x=0\end{array}\right.\) thỏa mãn đề bài
\(a,0.25x\) - \(\frac{2}{3}x\) =\(1\frac{1}{6}\)
\(\Rightarrow x\cdot\left(0.25-\frac{2}{3}\right)=1\frac{1}{6}\)
\(\Rightarrow x\cdot\frac{-5}{12}=\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{6}:\frac{-5}{12}\)
\(\Rightarrow x=\frac{-14}{5}=-2.8\)
Lời giải:
Đặt 3x−14=7y−45=t⇒x=4t+13;y=5t+473�−14=7�−45=�⇒�=4�+13;�=5�+47
Khi đó:
t=3x+7y−53x=4t+1+(5t+4)−54t+1�=3�+7�−53�=4�+1+(5�+4)−54�+1
⇒t=9t4t+1⇒�=9�4�+1
⇒t(4t+1)=9t⇒�(4�+1)=9�
⇒t(4t+1−9)=0⇒�(4�+1−9)=0
⇒t(4t−8)=0⇒�(4�−8)=0
⇒t=0⇒�=0 hoặc t=2�=2
Đến đây bạn thay vào tìm x,y thôi.