nhìn đã nản

e) \(\dfrac{x-5}{7}=\dfrac{x+2}{3}\Leftrightarrow3\left(x-5\right)=7\left(x+2\right)\Leftrightarrow3x-15=7x+14\)

\(\Leftrightarrow7x-3x=-15-14\Leftrightarrow4x=-29\Leftrightarrow x=\dfrac{-29}{4}\) vậy \(x=\dfrac{-29}{4}\)

f) \(\left(x-5\right)\left(x+2\right)=\left(x-3\right)\left(x+4\right)\Leftrightarrow x^2+2x-5x-10=x^2+4x-3x-12\)

\(\Leftrightarrow x^2+2x-5x-10-x^2-4x+3x+12=0\Leftrightarrow-4x+2=0\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\)

vậy \(x=\dfrac{1}{2}\)

a, \(2\dfrac{1}{3}:\dfrac{1}{3}=\dfrac{7}{9}:x\)

\(\Rightarrow\dfrac{7}{9}:x=\dfrac{7}{3}\cdot3\)

\(\Rightarrow\dfrac{7}{9}:x=7\)

\(\Rightarrow x=\dfrac{1}{9}\)

b, \(x:\dfrac{1}{3}=\dfrac{12}{99}:\dfrac{15}{90}\)

\(\Rightarrow x:\dfrac{1}{3}=\dfrac{4}{33}\cdot\dfrac{3}{18}\)

\(\Rightarrow x:\dfrac{1}{3}=\dfrac{2}{99}\)

\(\Rightarrow x=\dfrac{2}{297}\)

c, \(\dfrac{4}{9}:x=3\dfrac{1}{3}:2,25\)

\(\Rightarrow\dfrac{4}{9}:x=\dfrac{10}{3}\cdot\dfrac{4}{9}\)

\(\Rightarrow\dfrac{4}{9}:x=\dfrac{40}{27}\)

\(\Rightarrow x=\dfrac{3}{10}\)

d, \(\dfrac{4}{3}:\dfrac{41}{99}=x:\dfrac{75}{90}\)

\(\Rightarrow x:\dfrac{5}{6}=\dfrac{4}{3}\cdot\dfrac{99}{41}\)

\(\Rightarrow x:\dfrac{5}{6}=\dfrac{132}{41}\)

\(\Rightarrow x=\dfrac{110}{41}\)

e, \(\dfrac{x-5}{7}=\dfrac{x+2}{3}\)

\(\Rightarrow3\left(x-5\right)=7\left(x+2\right)\)

\(\Rightarrow3x-15=7x+14\)

\(\Rightarrow7x-3x=-15-14\)

\(\Rightarrow4x=-29\)

\(\Rightarrow x=-\dfrac{29}{4}\)

f, \(\left(x-5\right)\cdot\left(x+2\right)=\left(x-3\right)\cdot\left(x+4\right)\)

\(\Rightarrow x^2+2x-5x-10=x^2+4x-3x-12\)

\(\Rightarrow x^2-3x-10=x^2+x-12\)

\(\Rightarrow x^2-x^2-3x-x=-12+10\)

\(\Rightarrow-4x=2\)

\(\Rightarrow x=-0,5\)

sao trường khác học nhanh nhỉ? mk còn chưa học căn bậc 2 kìa, toàn lôi máy ra tính ko ak

Vch cái đề này 1 là bấm máy 2 là khai triển ra là xong

Tek cũng hỏi nx

\(A=\left(\sqrt{64}+2\sqrt{\left(-3\right)^2}-7\sqrt{1,69}+3\sqrt{\dfrac{25}{16}}\right):\left(5\sqrt{\dfrac{2}{3}}\right)^2\)

\(A=\left(\sqrt{64}+2\sqrt{9}-7\sqrt{\dfrac{169}{100}}+3\sqrt{\dfrac{25}{16}}\right):\left(5\sqrt{\dfrac{2}{3}}\right)^2\)

\(A=\left(8+2.3-7.\dfrac{13}{10}+3.\dfrac{5}{4}\right):\left(25.\dfrac{2}{3}\right)\)

\(A=\left(8+6-\dfrac{91}{10}+\dfrac{15}{4}\right):\left(\dfrac{50}{3}\right)\)

\(A=\dfrac{519}{1000}\)

Tương tự

Giải:

a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)

\(\Leftrightarrow24x-16-13x=60-15x+7x\)

\(\Leftrightarrow24x-13x+15x-7x=60+16\)

\(\Leftrightarrow19x=76\)

\(\Leftrightarrow x=\dfrac{76}{19}=4\)

Vậy ...

b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)

ĐKXĐ: \(x\ne\pm2\)

\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)

\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)

\(\Leftrightarrow8x^2-13x=0\)

\(\Leftrightarrow x\left(8x-13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)

Vậy ...

c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)

ĐKXĐ: \(x\ne-1;x\ne3\)

\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)

\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)

\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)

\(\Leftrightarrow x\left(2x-2\right)=4x\)

\(\Leftrightarrow2x-2=4\)

\(\Leftrightarrow x=3\)

Vậy ...

Mình chỉ làm những câu rõ đề thôi nhé ^^

1/ a/ Đặt \(t=2x-3\) thì pt trở thành \(t^3=\left(t+2\right)^2\Leftrightarrow t^3-t^2-4t-4=0\Leftrightarrow t^2\left(t-1\right)-4\left(t-1\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(t^2-4\right)=0\Leftrightarrow\left(t-2\right)\left(t-1\right)\left(t+2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}t=2\\t=1\\t=-2\end{array}\right.\)

Tới đây dễ rồi .

b/ Tương tự đặt \(a=2x-3\) thì pt trở thành \(a^3=a+2\Leftrightarrow a^3-a-2=0\)

Bạn xem lại đề , lớp 7 chưa học giải pt này đâu

c/ VT > 0 => VP > 0 => x > 0

Với x > 0 thì: \(\left|x+3\right|+\left|x+4\right|+\left|x+5\right|=x+3+x+4+x+5=3x+12\)

Tới đây dễ rồi :)

4) |2-|3-2x||=4

<=>\(\left[\begin{array}{nghiempt}2-\left|3-2x\right|=4\\2-\left|3-2x\right|=-4\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}\left|3-2x\right|=-2\left(vl\right)\\\left|3-2x\right|=6\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}3-2x=6\\3-2x=-6\end{array}\right.\)

<=> \(\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{9}{2}\end{array}\right.\)

\(\left|x+3\right|+\left|x+4\right|+\left|x+5\right|=4x\)

Ta có: \(\left|x+3\right|\ge0\)

\(\left|x+4\right|\ge0\)

\(\left|x+5\right|\ge0\)

\(\Rightarrow\left|x+3\right|+\left|x+4\right|+\left|x+5\right|\ge0\)

\(\Rightarrow4x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+3\right|+\left|x+4\right|+\left|x+5\right|=x+3+x+4+x+5=4x\)

\(\Rightarrow\left(x+x+x\right)+\left(3+4+5\right)=4x\)

\(\Rightarrow3x+12=4x\)

\(\Rightarrow x=12\)

Các câu còn lại đề sai hoặc t không biết làm ( thông cảm nhá!! )

a: \(A\left(x\right)+B\left(x\right)\)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)

\(=8x^2-12x\)

b: C(x)=A(x)-B(x)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)

\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)

Dài quá c ơi :<

a, \(\left(\dfrac{12}{25}\right)^x=\left(\dfrac{3}{5}\right)^2-\left(-\dfrac{3}{5}\right)^4\)

\(\Rightarrow\left(\dfrac{12}{25}\right)^x=\left(\dfrac{3}{5}\right)^2.\left[1-\left(\dfrac{3}{5}\right)^2\right]\)

\(\Rightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{9}{25}.\dfrac{16}{25}\Rightarrow\left(\dfrac{12}{25}\right)^x=\dfrac{144}{625}=\left(\dfrac{12}{25}\right)^2\)

Vì \(\dfrac{12}{25}\ne-1;\dfrac{12}{25}\ne0;\dfrac{12}{25}\ne1\) nên \(x=2\)

b, Sửa đề:

\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{81}{256}\)

\(\Rightarrow\left(-\dfrac{3}{4}\right)^{3x-1}=\left(-\dfrac{3}{4}\right)^4\)

Vì \(\dfrac{-3}{4}\ne-1;\dfrac{-3}{4}\ne0;\dfrac{-3}{4}\ne1\) nên \(3x-1=4\)

\(\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)

Vậy......

Chúc bạn học tốt!!!

Hồng Phúc Nguyễn, Hà Linh, Trương Hồng Hạnh, Hoàng Ngọc Anh, @Nguyễn Thanh Hằng, @Nguyễn Đinh Huyền Mai, ...

a: =2x^5-15x^3-x^2-2x^5-x^3=-16x^3-x^2

b: =x^3+3x^2-2x-3x^2-9x+6

=x^3-11x+6

c: \(=\dfrac{4x^3+2x^2-6x^2-3x-2x-1+5}{2x+1}\)

\(=2x^2-3x-1+\dfrac{5}{2x+1}\)

a) \(6x^3\left(\dfrac{1}{3}x^2-\dfrac{5}{2}-\dfrac{1}{6}\right)-2x^5-x^3\)

\(=6x^3\left(\dfrac{1}{3}x^2-\dfrac{16}{6}\right)-2x^5-x^3\)

\(=2x^5-16x^3-2x^5-x^3\)

\(=-17x^3\)

b) \(\left(x+3\right)\left(x^2+3x-2\right)\)

\(=x^3+3x^2-2x+3x^2+9x-6\)

\(=x^3+6x^2+7x-6\)

c) \(\left(4x^3-4x^2-5x+4\right):\left(2x+1\right)\)

\(=2x^2+4x^3-2x-4x^2-\dfrac{5}{2}-5x+\dfrac{2}{x}+4\)

\(=4x^3-2x^2-7x+\dfrac{2}{x}+\dfrac{3}{2}\)

1,

a,\(2x\left(3x^2-5x+3\right)\)

\(=6x^3-10x^2+6x\)

b,\(-2x\left(x^2+5x-3\right)\)

\(=-2x^3-10x^2+6x\)

c,\(-\dfrac{1}{2}x\left(2x^3-4x+3\right)\)

\(=-x^4+2x^2-\dfrac{3}{2}x\)

Bài 2:

a) \(\left(2x-1\right)\left(x^2-5-4\right)\)

\(=\left(2x-1\right)\left(x^2-9\right)\)

\(=2x^3-18x-x^2+9\)

b) \(-\left(5x-4\right)\left(2x+3\right)\)

\(=-\left(10x^2+15x-8x-12\right)\)

\(=-10x^2-7x+12\)

c) \(\left(2x-y\right)\left(4x^2-2xy+y^2\right)\)

\(=8x^3-y^3\)

Bài 3: (chỉ cần kết quả ko chứa biến là ta có đpcm, nói chung bài này yêu cầu ta rút gọn)

a) \(x\left(3x+12\right)-\left(7x-20\right)+x^2\left(2x-3\right)-x\left(2x^2+5\right)\)

\(=3x^2+12x-7x+20+2x^3-3x^2-2x^3-5x\)

\(=20\)

b) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)-19x\)

\(=6x-3-5x+15+18x-24-19x\)

\(=-12\)