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a: \(x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b: \(2a\left(x+y\right)+x+y=\left(x+y\right)\left(2a+1\right)\)
c: \(6x^2-12x-7x+14\)
\(=6x\left(x-2\right)-7\left(x-2\right)\)
\(=\left(x-2\right)\left(6x-7\right)\)
\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)
\(10x-15-20x+28=19-2x-22\)
\(10x-20x+2x=19-22-28+15\)
\(-8x=-16\)
\(\Rightarrow x=2\)
\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)
\(4x+12-7x+17=40x-8+166\)
\(4x-7x-40x=-8+166-17-12\)
\(-43x=129\)
\(x=-3\)
\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)
\(17-14x+14=13-4x-4-5x+15\)
\(-14x+4x+5x=13-4+15-14-17\)
\(-5x=-7\)
\(x=\frac{7}{5}\)
\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)
\(5x+3,5+3x-4=7x-3x+1,5\)
\(5x+3x-7x+3x=1,5-3,5\)
\(x=-2\)
\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)
\(28x+21-4x+4=15x+11,25+7\)
\(28x-4x-15x=11,25+7-4-21\)
\(9x=\frac{-27}{4}\)
\(x=\frac{-3}{4}\)
\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)
\(3x+0,8x+0,2x=3,38-2,42\)
\(4x=\frac{24}{25}\)
\(x=\frac{6}{25}\)
chúc bạn học tốt !!
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a: \(\dfrac{x^2-5x+6}{x^2+7x+12}:\dfrac{x^2-4x+4}{x^2+3x}\)
\(=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+3\right)\left(x+4\right)}\cdot\dfrac{x\left(x+3\right)}{\left(x-2\right)^2}\)
\(=\dfrac{x\left(x-3\right)}{\left(x-2\right)\left(x+4\right)}\)
b: \(\dfrac{x^2+2x-3}{x^2+3x-10}:\dfrac{x^2+7x+12}{x^2-9x+14}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x+4\right)}\)
\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x+4\right)}\)
Bài 1:
a) Ta có: 7x+12=0
\(\Leftrightarrow7x=-12\)
hay \(x=-\frac{12}{7}\)
Vậy: \(x=-\frac{12}{7}\)
b) Ta có: 5x-2=0
\(\Leftrightarrow5x=2\)
hay \(x=\frac{2}{5}\)
Vậy: \(x=\frac{2}{5}\)
c) Ta có: 12-6x=0
\(\Leftrightarrow6x=12\)
hay x=2
Vậy: x=2
d) Ta có: -2x+14=0
⇔-2x=-14
hay x=7
Vậy: x=7
Bài 2:
a) Ta có: 3x+1=7x-11
⇔3x+1-7x+11=0
⇔-4x+12=0
⇔-4x=-12
hay x=3
Vậy: x=3
b) Ta có: 2x+x+12=0
⇔3x+12=0
⇔3x=-12
hay x=-4
Vậy: x=-4
c) Ta có: x-5=3-x
⇔x-5-3+x=0
⇔2x-8=0
⇔2x=8
hay x=4
Vậy: x=4
d) Ta có: 7-3x=9-x
⇔7-3x-9+x=0
⇔-2x-2=0
⇔-2x=2
hay x=-1
Vậy: x=-1
e) Ta có: 5-3x=6x+7
⇔5-3x-6x-7=0
⇔-9x-2=0
⇔-9x=2
hay \(x=\frac{-2}{9}\)
Vậy: \(x=\frac{-2}{9}\)
f) Ta có: 11-2x=x-1
⇔11-2x-x+1=0
⇔12-3x=0
⇔3x=12
hay x=4
Vậy: x=4
g) Ta có: 15-8x=9-5
⇔15-8x=4
⇔8x=11
hay \(x=\frac{11}{8}\)
Vậy: \(x=\frac{11}{8}\)
Bài 3:
a) Ta có: 0,25x+1,5=0
⇔0,25x=-1,5
hay x=-6
Vậy: x=-6
b) Ta có: 6,36-5,2x=0
⇔5,2x=6,36
hay \(x=\frac{159}{130}\)
Vậy: \(x=\frac{159}{130}\)
Ta có phép chia
Chọn đáp án B.