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1/ Ta có: \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
\(\Rightarrow5x^2-15x=5x^2-11x+2-5\)
\(\Rightarrow-15x=-11x-3\)
\(\Rightarrow-15x+11x=-3\)
\(\Rightarrow-4x=-3\Rightarrow x=\frac{3}{4}\).
2/ Ta có: \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
\(\Rightarrow2x^2-5x+2-2x^2+x+21-3=0\)
\(\Rightarrow-4x+20=0\Rightarrow-4x=-20\Rightarrow x=5\).
Chúc bn hc tốt!
1/ \(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\)
<=> \(5x^2-15x=\left(5x^2-x-10x+2\right)-5\)
<=> \(5x^2-15x=5x^2-11x-3\)
<=> \(5x^2-15x-5x^2+11x+3=0\)
<=> \(-4x+3=0\)
<=> \(x=\frac{3}{4}\)
2/ \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
<=> \(2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)
<=> \(2x^2-5x+2-2x^2+x+21=3\)
<=> \(-4x+21=3\)
<=> \(4x=17\)
<=> \(x=\frac{17}{4}\)
+) (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
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
Không có đề bài thì mình chịu!
a) \(\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\)
\(\Leftrightarrow x+1=\frac{\left(2x-1\right)\left(x+5\right)}{2x-3}\)
\(\Leftrightarrow x=\frac{\left(2x-1\right)\left(x+5\right)}{2x-3}-1\)
b) \(\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
\(\Leftrightarrow\)\(\left(x-1\right)^3=5x\left(2-x\right)-11\left(x+2\right)+x\left(x+1\right)^2\)
\(\Leftrightarrow x-1=\sqrt[3]{5x\left(2-x\right)-11\left(x+2\right)+x\left(x+1\right)^2}\)
\(\Leftrightarrow x=\sqrt[3]{5x\left(2-x\right)-11\left(x+2\right)+x\left(x+1\right)^2}+1\)
Dạng ax+b=0