a) \(=\frac{25}{36}\)

b) \(=\frac{50}{7}\)

c) \(=\frac{200}{279}\)

a) 1,2: 3,24 = 120 : 324 = 10:27

b) 215:34215:34 = 115:34=115.43=44:15115:34=115.43=44:15

c) 27:0,4227:0,42 = 27:42100=27.10042=200294=100147=100:14727:42100=27.10042=200294=100147=100:147

\(x^2\)\(+x-2022x-2022=0\)

\(x\left(x-2022\right)\)\(+\left(x-2022\right)\)\(=0\)

\(\left(x-2022\right)\)\(\left(x+1\right)\)\(=0\)

\(=>TH1:x-2022=0\)

\(=>x=2022\)

\(TH2:x-1=0\)

\(=>x=-1\)

Đặt A(x)=x2−2021x+2020=0A(x)=x2−2021x+2020=0

⇔x2−2020x−x+2020=0⇔x2−2020x−x+2020=0

⇔x(x−1)−2020(x−1)=0⇔x(x−1)−2020(x−1)=0

⇔(x−2020)(x−1)=0⇔x=\orbr{x=2020x=1⇔(x−2020)(x−1)=0⇔x=\orbr{x=2020x=1

Vậy nghiệm của phương trình là x = 1 ; x = 2020 

Ta có :

\(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

\(\Rightarrow\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}-\frac{x+2}{14}-\frac{x+2}{15}=0\)

\(\Rightarrow\left(x+2\right).\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\right)=0\)

\(\Rightarrow x+2=0\left(\text{do }\frac{1}{11}+\frac{1}{12}+\frac{1}{13}-\frac{1}{14}-\frac{1}{15}\ne0\right)\)

\(\Rightarrow x=-2\)

Bài 2 ; 2 ; 3 ; 4 phần tự luận bạn tham khảo ở Câu hỏi của Phùng minh long - Toán lớp 7 - Học trực tuyến OLM nhé !

Bài 1 :

\(1)\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}\)

\(\Rightarrow\left(\frac{11}{24}+\frac{13}{24}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)+\frac{1}{2}\)

\(\Rightarrow1-1+\frac{1}{2}=\frac{1}{2}\)

\(2)\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1\frac{15}{17}+\frac{2}{3}\)

\(=\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-1-\frac{15}{17}+\frac{14}{21}\)

\(=\left(\frac{15}{34}+\frac{19}{34}-1\right)+\left(\frac{7}{21}+\frac{14}{21}\right)-\frac{15}{17}\)

\(=1-\frac{15}{17}=\frac{2}{17}\)

\(3)1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)

\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)

\(=1+1+0,5=2,5\)

\(4)\left(-12\right):\left(\frac{3}{4}-\frac{5}{6}\right)^2\)

\(=\left(-12\right):\left(\frac{-1}{12}\right)^2\)

\(=\left(-12\right).12^2=-12^3\)

\(5)\left(\frac{9}{25}-2.18\right):\left(3\frac{4}{5}+0,2\right)\)

\(=\frac{-891}{25}:4=\frac{-891}{100}\)

\(6)\left(-6,5\right).5,7+5,7.\left(-3,5\right)\)

\(=\left[\left(-6,5\right)+\left(-3,5\right)\right].5,7\)

\(=-10.5,7=57\)

\(7)-3,75.\left(-7,2\right)+2,8.3,75\)

\(=3,75.7,2+2,8.3,75\)

\(=3,75.\left(7,2+2,8\right)\)

\(=3,75.10=37,5\)

\(8)23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}:\frac{5}{7}\)

\(=23\frac{1}{4}.\frac{7}{5}-13\frac{1}{4}.\frac{7}{5}\)

\(=\left(23\frac{1}{4}-13\frac{1}{4}\right).1,4\)

\(=10.1,4=14\)

\(9)16\frac{2}{7}:\left(\frac{-3}{5}\right)+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}:\frac{3}{5}+28\frac{2}{7}:\frac{3}{5}\)

\(=-16\frac{2}{7}.\frac{5}{3}+28\frac{2}{7}.\frac{5}{3}\)

\(=\left(-16\frac{2}{7}+28\frac{2}{7}\right).\frac{5}{3}\)

\(=12.\frac{5}{3}=20\)

    3456789123456789

Bài 2 :

\(\left(2x+1\right)^2=\frac{16}{25}\)

\(\Rightarrow\left(2x+1\right)^2=\left(\frac{4}{5}\right)^2=\left(-\frac{4}{5}\right)^2\)

\(\Rightarrow\orbr{\begin{cases}2x+1=\frac{4}{5}\\2x+1=-\frac{4}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{-1}{5}\\2x=-\frac{9}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{10}\\x=-\frac{9}{10}\end{cases}}\)

b) \(0,5-\left|\frac{5}{4}-2x\right|=\frac{1}{3}\)

\(\Rightarrow\frac{1}{2}-\left|\frac{5}{4}-2x\right|=\frac{1}{3}\)

\(\Rightarrow\left|\frac{5}{4}-2x\right|=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)

\(\Rightarrow\orbr{\begin{cases}\frac{5}{4}-2x=\frac{1}{6}\\\frac{5}{4}-2x=\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{5}{4}-\frac{1}{6}\\2x=\frac{5}{4}-\frac{-1}{6}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{13}{12}\\2x=\frac{17}{12}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{24}\\x=\frac{17}{24}\end{cases}}\)

c) Ta có :

\(\frac{x}{-3}=\frac{y}{7}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\frac{x}{-3}=\frac{y}{7}=\left(\frac{x}{-3}\right)^2=\left(\frac{y}{7}\right)^2=\frac{x^2}{\left(-3\right)^2}=\frac{y^2}{7^2}=\frac{x^2-y^2}{\left(-3\right)^2-7^2}=\frac{-160}{40}=4\)

\(\Rightarrow\hept{\begin{cases}x=-3.4=-12\\y=7.4=28\end{cases}}\)

Bài  5 :

Ta có : a + c = 2b và 2bd = c . ( b + d ) ( b và d khác 0 )

=> ( a + c ) . d = c . ( b + d )

=> ad + cd = cb + cd

=> ad = cb

\(\Rightarrow\frac{ad}{bd}=\frac{cb}{bd}\Rightarrow\frac{a}{b}=\frac{c}{b}\)

bài nào vậy bn

BÀI NÀO

Cách 1: Vẽ đường thằng x’x . Trên x’x lấy điểm O

Vẽ góc  (Oy và Oy’ cùng thuộc một nửa mặt phẳng bờ xx’).

~ Chúc bạn hok tốt ~ ^^

Mình tí giải tiếp cho bạn nhé,mình còn đang làm bài bồi toán(thông cảm)

1)11/24-5/41+13/24+0,5-36/41=(11/24+13/24)-(5/41+36/41)+0,5=24/24-41/41+0,5=1-1+0,5=0,5

2)15/34+7/21+19/34-1​​và 15/17+2/3=(15/34+19/34)+(1/3+2/3)-32/17=34/34+3/3-32/17=1+1/32/17=2+32/17=34/17+32/17=66/17

3)1 và 4/23+5/21-4/23+0,5+16/21=(27/23-4/23)+(5/21+16/21)+0,5=23/23+21/21+0,5=1+1+0,5=2,5

4)(-12):(3/4-5/6)²=(-12):(9/12+10/12)²=(-12):(19/12)²=(-12):19²/12²=(-19)²/12=361/12

a) \(3y\left(2y-4\right)-2y\left(3y+5\right)=44\)

\(\Rightarrow6y^2-12y-6y^2-10y=44\)

\(\Rightarrow\left(6y^2-6y^2\right)-\left(12y+10y\right)=44\)

\(\Rightarrow-22y=44\)

\(\Rightarrow y=-2\)

b) \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)

\(\Rightarrow48x^2-12x-20x+5+3x-48x^2-7+112x=81\)

\(\Rightarrow83x-2=81\)

\(\Rightarrow x=1\)

a) 3^4-x = 3³

\(\Rightarrow\)4 - x = 3

              x = 4 - 3

              x = 1

b) ( x - \(\frac{4}{5}\))² = 0²

\(\Rightarrow\)x - \(\frac{4}{5}=0\)

         x         = 0 + \(\frac{4}{5}\)

         x         = \(\frac{4}{5}\)

a, đề bài

---> \(3^{4-x}=3^3\)

---> 4 - x = 3

---> x = 1

b, đề bài

---> \(x-\frac{4}{5}=0\)

---> \(x=\frac{4}{5}\)

xin tiick