a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

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

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

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

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)

a) \(64x^3-\dfrac{1}{8}y^3=\left(4x-\dfrac{1}{2}y\right)\left(16x^2+2xy+\dfrac{1}{4}y^2\right)\)

b) \(125x^6-27x^9=\left(5x^2-3x^3\right)\left(25x^4+15x^5+9x^6\right)\)

c) \(-\dfrac{x^6}{125}-\dfrac{y^3}{64}=-\left(\dfrac{x^6}{125}+\dfrac{y^3}{64}\right)=-\left(\dfrac{x^2}{5}+\dfrac{y}{4}\right)\left(\dfrac{x^4}{25}-\dfrac{x^2y}{20}+\dfrac{y^2}{16}\right)\)

Các câu sai: a, c, d, f

Các câu đúng: b, e

Các câu đúng: b,e
Các câu sai: a, c, d; f.
a) \(\left(-5\right)^2.\left(-5\right)^3=\left(-5\right)^5\);
c) \(\left(0,2\right)^{10}:\left(0,2\right)^5=\left(0,2\right)^{10-5}=0,2^5\);
d) \(\left[\left(-\dfrac{1}{7}\right)^2\right]^4=\left(-\dfrac{1}{7}\right)^{2.4}=\left(-\dfrac{1}{7}\right)^8\)
f \(\dfrac{8^{10}}{4^8}=\dfrac{\left(2^3\right)^5}{\left(2^2\right)^8}=\dfrac{2^{15}}{2^{16}}=\dfrac{1}{2}\)

- Các câu sai là a, c, d, f

- Các câu đúng là b, e

Sửa lại:

a) (-5)2 .(-5)3 = (-5)5

c) (0,2)10 : (0,2)5 = (0,2)5

\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)

\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)

\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)

\(A=\dfrac{1}{2}-\dfrac{1}{4}\)

\(A=\dfrac{2}{4}-\dfrac{1}{4}\)

\(A=\dfrac{1}{4}\)

\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)

\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)

\(D=\dfrac{77}{39}+\dfrac{3}{2}\)

\(D=\dfrac{271}{78}\)

\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)

\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)

\(C=\dfrac{5}{2}-\dfrac{3}{2}\)

\(C=1\)

a: Ta có: \(A=\sqrt{8}-2\sqrt{18}+3\sqrt{50}\)

\(=2\sqrt{2}-6\sqrt{2}+15\sqrt{2}\)

\(=11\sqrt{2}\)

b: Ta có: \(B=\sqrt{125}-10\sqrt{\dfrac{1}{20}}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)

\(=5\sqrt{5}-\sqrt{5}+\sqrt{5}-1\)

\(=5\sqrt{5}-1\)

\(a,-\dfrac{3}{5}-x=-0,75\\ -\dfrac{3}{5}-x=-\dfrac{3}{4}\\ x=-\dfrac{3}{5}-\left(-\dfrac{3}{4}\right)\\ x=-\dfrac{3}{5}+\dfrac{3}{4}=\dfrac{3}{20}\\ ---\\ b,1\dfrac{4}{5}=-0,15-x\\ \dfrac{9}{5}=-\dfrac{3}{20}-x\\ x=-\dfrac{3}{20}-\dfrac{9}{5}\\ x=-\dfrac{3}{20}-\dfrac{36}{20}\\ x=-\dfrac{39}{20}\\ ----\\ c,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}+\dfrac{4}{7}\\ \dfrac{33}{10}-x=\dfrac{26}{21}\\ x=\dfrac{33}{10}-\dfrac{26}{21}\\ x=\dfrac{433}{210}\)

a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)

b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)

c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)

a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) =  \(\dfrac{6}{7}\)  

b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)          

c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)

a) \(-\dfrac{3}{5}-x=-0,75\)

\(\Rightarrow-\dfrac{3}{5}-x=-\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{3}{5}\)

\(\Rightarrow x=\dfrac{15}{20}-\dfrac{12}{20}=\dfrac{8}{20}=\dfrac{2}{5}\)

b) \(1\dfrac{4}{5}=-0,15-x\)

\(\Rightarrow\dfrac{9}{5}=-\dfrac{3}{20}-x\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{9}{5}\)

\(\Rightarrow x=-\dfrac{3}{20}-\dfrac{36}{20}=-\dfrac{39}{20}\)

c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)\)

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

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

a) \(-\dfrac{3}{5}-x=-0,75\)

\(x=-\dfrac{3}{5}+0,75=\dfrac{3}{5}+\dfrac{3}{4}\)

\(x=\dfrac{27}{20}\)

________

b) \(1\dfrac{4}{5}=-0,15-x\)

\(=>-0,15-x=\dfrac{9}{5}\)

\(x=\dfrac{-3}{20}-\dfrac{9}{5}=\dfrac{-3}{20}-\dfrac{36}{20}\)

\(x=\dfrac{-39}{20}\)

c) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\left(-\dfrac{1}{3}\right)=\dfrac{6}{15}+\dfrac{5}{15}\)

\(x+\dfrac{1}{3}=\dfrac{11}{15}\)

\(x=\dfrac{11}{15}-\dfrac{1}{3}=\dfrac{11}{15}-\dfrac{5}{15}\)

\(x=\dfrac{6}{15}=\dfrac{2}{5}\)

Mik sửa lại câu a

\(...x=-\dfrac{3}{5}--0,75=\dfrac{-3}{5}+\dfrac{3}{4}\)

\(x=\dfrac{-12}{20}+\dfrac{15}{20}\)

\(x=\dfrac{3}{20}\)

Viết hết tất cả dưới dạng tự luận luôn à bạn?

Câu 2: C

C1. Mik ko bik vì bấm máy tính thì nó ra kết quả quá lớn.

C2. \(\dfrac{15}{19}.\dfrac{2}{3}-\dfrac{7}{19}.\dfrac{2}{3}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\left(\dfrac{15}{19}-\dfrac{7}{19}\right)+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{2}{3}.\dfrac{8}{19}+\dfrac{8}{3}.\dfrac{17}{19}\\ =\dfrac{16}{57}+\dfrac{136}{57}\\ =\dfrac{152}{57}\\ =\dfrac{8}{3}\left(C\right)\)

C3. \(x=4\left(B\right)\)

C4. Gọi tổng số tiền ba đợt ủng hộ lần lượt là a, b, c (a, b, c ϵ N*).

Vì số tiền ủng hộ ba đợt lần lượt tỉ lệ với 7; 8; 9

Nên \(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}\)

Vì tổng số tiền ủng hộ đợt hai và đợt ba nhiều hơn số tiền ủng hộ đợt một 80 tỉ

Nên \(\left(b+c\right)-a=80\)

Theo tính chất của dãy tỉ số bằng nhau

Ta có:

\(\dfrac{a}{7}=\dfrac{b}{8}=\dfrac{c}{9}=\dfrac{\left(b+c\right)-a}{\left(8+9\right)-7}=\dfrac{80}{10}=8\)

Do đó:

\(\dfrac{a}{7}=8=>a=8.7=>a=56\\ \dfrac{b}{8}=8=>b=8.8=>b=64\\ \dfrac{c}{9}=8=>c=8.9=>c=72\)

Vậy số tiền ủng hộ ba đợt lần lượt là: 56; 64; 72.

Số tiền ủng hộ đợt hai là 64 tỉ (B).

Các đa thức 1 biến là :

\(3 + 6y;7{x^2} + 2x - 4{x^4} + 1;\dfrac{1}{3}x - 5\)