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\(1,\frac{x+1}{x-2}=\frac{3}{4}\)
\(\Rightarrow3x-6=4x+4\)
\(\Rightarrow3x-4x=4+6\)
\(\Rightarrow-x=10\Leftrightarrow x=-10\)
\(2,\frac{x-1}{3}=\frac{x+3}{5}\)
\(\Rightarrow5x-5=3x+9\)
\(\Rightarrow5x-3x=9+5\)
\(\Rightarrow2x=14\Leftrightarrow x=7\)
\(3,\frac{2x+3}{24}=\frac{3x-1}{32}\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow72x-64x=24+96\)
\(\Rightarrow8x=120\)
\(\Rightarrow x=15\)
a: \(\Leftrightarrow x\cdot\dfrac{2}{7}-\dfrac{3}{5}x=-1+\dfrac{1}{3}\)
\(\Leftrightarrow x\cdot\dfrac{-11}{35}=\dfrac{-2}{3}\)
hay x=70/33
b: \(\left|3x+7\right|-3x=7\)
\(\Leftrightarrow\left|3x+7\right|=3x+7\)
\(\Leftrightarrow3x+7\ge0\)
hay x>=-7/3
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
\(2A\left(x\right)-B\left(x\right)=2\left(-3x^4+3x^3+7x^2-6x-2\right)-\left(-5x^4+2x^3-x^2+7\right)\)
\(=-6x^4+6x^3+14x^2-12x-4+5x^4-2x^3+x^2-7\)
\(\Rightarrow2A\left(x\right)-B\left(x\right)=-x^4+4x^3+15x^2-12x-11\)
Thu gọn đa thức được:
\(2x^5-4x^4-3x^2+1\)
Vậy bậc của đa thức là 5.(Nếu sai do tính toán thì đừng cmr nha)
dù sao thì cũng nên l i k e bài nào phải làm đủ hơn chứ vì đã nói là chỉ viết đấp số ngay từ đầu đâu
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
Ta có:
\(3x=2y\Rightarrow\frac{x}{2}=\frac{y}{3}.\) (1)
\(7x=5z\Rightarrow\frac{x}{5}=\frac{z}{7}.\) (2)
Từ (1) và (2) => \(\frac{x}{2}=\frac{y}{3};\frac{x}{5}=\frac{z}{7}\)
Có: \(\frac{x}{2}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{15}.\)
\(\frac{x}{5}=\frac{z}{7}\Rightarrow\frac{x}{10}=\frac{z}{14}.\)
\(\Rightarrow\frac{x}{10}=\frac{y}{15}=\frac{z}{14}\) và \(x-y+z=32.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{14}=\frac{x-y+z}{10-15+14}=\frac{32}{9}.\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{10}=\frac{32}{9}\Rightarrow x=\frac{32}{9}.10=\frac{320}{9}\\\frac{y}{15}=\frac{32}{9}\Rightarrow y=\frac{32}{9}.15=\frac{160}{3}\\\frac{z}{14}=\frac{32}{9}\Rightarrow z=\frac{32}{9}.14=\frac{448}{9}\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(\frac{320}{9};\frac{160}{3};\frac{448}{9}\right).\)
Chúc bạn học tốt!
a) 6x(5x + 3) + 3x(1 – 10x) = 7
⇒ 30x2+18x+3x-30x2=7
⇒21x=7
⇒x=\(\dfrac{7}{21}\)
⇒x= \(\dfrac{1}{3}\)
b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44
⇒15x-63x2-15+63x + 63x2-35x+36x-20=44
⇒79x-35=44
⇒79x=44+35
⇒79x=79
⇒x=1
2.3x + 3x - 1 = 7 . (32 + 2 . 62)
=> 2.3x + 3x - 1 = 567
=> 7 . 3x - 1 = 567
=> 3x - 1 = 567 : 7 = 81
=> x - 1 = 4
=> x = 5
a)2*3x+3x-1=7(32+2*62)
2*3x+3x-1=7(9+72)=7*81
2*3x+3x/3=567
2*3x+3x*1/3=567
(2+1/3)*3x=567
7/3*3x=567
3x=567:7/3
3x=243=35
=>x=5
b) mk ko hiểu đề mấy, cái chỗ 7x+2 là nhân vs 2 ak