a) x (x+1) + x (x-3) = 2.2x b) 90 (x-6) – 36x = 2x (x-6) 04/10/2021 Bởi Ayla a) x (x+1) + x (x-3) = 2.2x b) 90 (x-6) – 36x = 2x (x-6)
`\text{~~Holi~~}` `a. x(x+1)+x(x+3)=2.2x` `-> x^2+x+x^2+3x=4x` `-> 2x^2+4x=4x` `-> 2x^2=0` `-> x=0` Vậy `S={0}` `b. 90(x-6)-36x=2x(x-6)` `-> 90x-540-36x=2x^2-12x` `-> 54x-540=2x^2-12x` `-> 54x-540-2x^2+12x=0` `-> 66x-540-2x^2=0` `-> -2x^2+66x-540=0` `-> 2x^2-66x+540=0` `-> (x-18)(x-15)=0` `->`\(\left[ \begin{array}{l}x=18\\x=15\end{array} \right.\) Vậy `S={18;15}` Bình luận
Đáp án: Giải thích các bước giải: b,ta có: 90(x-6)-36x=2x”(x-6) <=>90x-540=2$x^{2}$-12x <=>66????−540−2????^2=0 <=>−2(????^2−33????+270)=0 <=>2(????^2−15????−18????+270)=0 <=>−2(????−18)(????−15)=0 <=>x=18;15 a,ta có: a. x(x+1)+x(x+3)=2.2x <=>x²+x+x²+3x-2.2x=0 <=>2x²+1.8x=0 <=> x(2x+1.8)=0 <=>x=0; x=-0.9 Bình luận
`\text{~~Holi~~}`
`a. x(x+1)+x(x+3)=2.2x`
`-> x^2+x+x^2+3x=4x`
`-> 2x^2+4x=4x`
`-> 2x^2=0`
`-> x=0`
Vậy `S={0}`
`b. 90(x-6)-36x=2x(x-6)`
`-> 90x-540-36x=2x^2-12x`
`-> 54x-540=2x^2-12x`
`-> 54x-540-2x^2+12x=0`
`-> 66x-540-2x^2=0`
`-> -2x^2+66x-540=0`
`-> 2x^2-66x+540=0`
`-> (x-18)(x-15)=0`
`->`\(\left[ \begin{array}{l}x=18\\x=15\end{array} \right.\)
Vậy `S={18;15}`
Đáp án:
Giải thích các bước giải:
b,ta có:
90(x-6)-36x=2x”(x-6)
<=>90x-540=2$x^{2}$-12x
<=>66????−540−2????^2=0
<=>−2(????^2−33????+270)=0
<=>2(????^2−15????−18????+270)=0
<=>−2(????−18)(????−15)=0
<=>x=18;15
a,ta có:
a. x(x+1)+x(x+3)=2.2x
<=>x²+x+x²+3x-2.2x=0
<=>2x²+1.8x=0
<=> x(2x+1.8)=0
<=>x=0;
x=-0.9