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(4x – 10)(24 + 5x) = 0 ⇔ 4x – 10 = 0 hoặc 24 + 5x = 0
4x – 10 = 0 ⇔ 4x = 10 ⇔ x = 2,5
24 + 5x = 0 ⇔ 5x = -24 ⇔ x = -4,8
Phương trình có nghiệm x = 2,5 và x = -4,8
a. (4x−10)(24+5x)=0⇔4x−10=0(4x−10)(24+5x)=0⇔4x−10=0 hoặc 24+5x=024+5x=0
+ 4x−10=0⇔4x=10⇔x=2,54x−10=0⇔4x=10⇔x=2,5
+ 24+5x=0⇔5x=24⇔x=−4,824+5x=0⇔5x=24⇔x=−4,8
Phương trình có nghiệm x = 2,5 và x = -4,8
b. (3,5−7x)(0,1x+2,3)=0⇔3,5−7x=0(3,5−7x)(0,1x+2,3)=0⇔3,5−7x=0hoặc 0,1x+2,3=00,1x+2,3=0
+ 3,5−7x=0⇔3,5=7x⇔x=0,53,5−7x=0⇔3,5=7x⇔x=0,5
+ 0,1x+2,3=0⇔0,1x=−2,3⇔x=−230,1x+2,3=0⇔0,1x=−2,3⇔x=−23
Phương trình có nghiệm x =0,5 hoặc x = -23
\(\left(x^2+5x^2\right)-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow4x^2-10x-24=0\)
\(\Leftrightarrow\frac{-\left(-10\right)+\sqrt{\left(-10\right)^2-4.4.\left(-24\right)}}{2.4}\)
\(\Leftrightarrow\frac{10+\sqrt{484}}{2.4}\)
\(\Leftrightarrow\frac{10+\sqrt{484}}{8}\)
\(\Leftrightarrow\frac{-\left(-10\right)-\sqrt{\left(-10\right)^2-4.4.\left(-24\right)}}{2.4}\)
\(\Leftrightarrow\frac{10-\sqrt{\left(10\right)^2+4.4.24}}{2.4}\)
\(\Leftrightarrow\frac{10-\sqrt{484}}{8}\)
\(\Rightarrow\hept{\begin{cases}x=4\\x=-\frac{3}{2}\end{cases}}\)
Sai đâu sửa hộ :)
Câu 1:
a) Ta có: 7x+21=0
\(\Leftrightarrow7x=-21\)
hay x=-3
Vậy: S={-3}
b) Ta có: 3x-2=2x-3
\(\Leftrightarrow3x-2-2x+3=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Vậy: S={-1}
c) Ta có: 5x-2x-24=0
\(\Leftrightarrow3x=24\)
hay x=8
Vậy: S={8}
Câu 2:
a) Ta có: \(\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{1}{2};1\right\}\)
b) Ta có: \(\left(2x-3\right)\left(-x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\-x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\-x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=7\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{3}{2};7\right\}\)
c) Ta có: \(\left(x+3\right)^3-9\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)^2-9\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+3-3\right)\left(x+3+3\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=-6\end{matrix}\right.\)
Vậy: S={0;-3;-6}
\(\left(x^2+5x\right)^2-2x^2-10x=24\)
\(\Leftrightarrow\left[x\left(x+5\right)\right]^2-2x\left(x+5\right)-24=0\)
\(\Leftrightarrow\left[x\left(x+5\right)\right]^2-2x\left(x+5\right)+1-25=0\)
\(\Leftrightarrow\left[x\left(x+5\right)-1\right]^2-5^2=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)\left(x+1\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\Leftrightarrow x=1\\x+6=0\Leftrightarrow x=-6\\x+1=0\Leftrightarrow x=-1\\x+4=0\Leftrightarrow x=-4\end{matrix}\right.\)
\(\left|x^2-5x-6\right|=x^2-x-24\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-6=x^2-x-24\\x^2-5x-6=x-x^2+24\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-4x+18=0\\2x^2-6x-30=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(loai\right)\\\left[{}\begin{matrix}x=\dfrac{3+\sqrt{69}}{2}\left(tm\right)\\x=\dfrac{3-\sqrt{69}}{2}\left(loai\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy ...
3x+24=5x-60
3x - 5x = -60 - 24
x ( 3-5) = -84
x (-2) = -84
x = -84 : ( -2)
x = 42
Theo cách giải 6
3x+24=5x-60
<=> 24+60=5x-3x
<=> 84=2x
<=> x=42
Vậy phương trình có nghiệm x=42