(X+1)(2x-3)=(2x-1)(x+5).
(X-1)^3-x(x+1)^2=5x(2-x)-11(x+2). Mình cần gấp mong các bạn giúp
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NT
3
12 tháng 6 2018
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!
HH
12 tháng 6 2018
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/ \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)
<=> \(2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)
<=> \(2x^2-5x+2-2x^2+x+21=3\)
<=> \(-4x+21=3\)
<=> \(4x=17\)
<=> \(x=\frac{17}{4}\)
12 tháng 6 2018
+) (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
29 tháng 6 2019
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
29 tháng 6 2019
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
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