1/9+(x-1/3)²=5/6
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\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)
\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)
\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)
\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)
\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)
\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)
\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)
\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)
\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)
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`@` `\text {Ans}`
`\downarrow`
`1)`
\(x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow` \(x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{7}{6}\)
Vậy, `x =`\(\dfrac{7}{6}\)
`2)`
\(\dfrac{3}{5}-x=\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{3}{5}-\dfrac{1}{3}\)
`\Rightarrow`\(x=\dfrac{4}{15}\)
Vậy, `x =`\(\dfrac{4}{15}\)
`3)`
\(\dfrac{3}{4}+x=\dfrac{7}{2}\)
`\Rightarrow`\(x=\dfrac{7}{2}-\dfrac{3}{4}\)
`\Rightarrow`\(x=\dfrac{11}{4}\)
Vậy, \(x=\dfrac{11}{4}\)
`4)`
\(x-\dfrac{4}{3}=\dfrac{7}{9}\)
`\Rightarrow`\(x=\dfrac{7}{9}+\dfrac{4}{3}\)
`\Rightarrow`\(x=\dfrac{19}{9}\)
Vậy, `x=`\(\dfrac{19}{9}\)
`5)`
\(x-\dfrac{5}{6}=\dfrac{7}{3}\)
`\Rightarrow`\(x=\dfrac{7}{3}+\dfrac{5}{6}\)
`\Rightarrow x =`\(\dfrac{19}{6}\)
Vậy, `x=`\(\dfrac{19}{6}\)
`6)`
\(x-\dfrac{1}{5}=\dfrac{9}{10}\)
`\Rightarrow x=`\(\dfrac{9}{10}+\dfrac{1}{5}\)
`\Rightarrow x=`\(\dfrac{11}{10}\)
Vậy, `x=`\(\dfrac{11}{10}\)

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)

1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)
\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)
\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)
Vậy \(x=\dfrac{-13}{6}\)
2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)
\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)
\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)
\(-x=\dfrac{4}{15}\)
Vậy \(x=\dfrac{-4}{15}\)
3/ \(3-4+x=\dfrac{7}{2}\)
\(-4+x=\dfrac{7}{2}-3\)
\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)
\(-4+x=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}+4\)
\(x=\dfrac{1}{2}+\dfrac{8}{2}\)
Vậy \(x=\dfrac{9}{2}\)
4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)
\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)
\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)
Vậy \(x=\dfrac{5}{9}\)
5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)
\(x=\dfrac{5}{6}-\dfrac{7}{2}\)
\(x=\dfrac{5}{6}-\dfrac{21}{6}\)
Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)
6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)
\(x=\dfrac{9}{10}+\dfrac{1}{5}\)
\(x=\dfrac{9}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{11}{10}\)
7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)
\(x=\dfrac{3}{8}-\dfrac{5}{12}\)
\(x=\dfrac{9}{24}-\dfrac{10}{24}\)
Vậy \(x=\dfrac{-1}{24}\)
8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)
\(x=\dfrac{7}{6}-\dfrac{5}{4}\)
\(x=\dfrac{14}{12}-\dfrac{15}{12}\)
Vậy \(x=\dfrac{-1}{12}\)
9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)
\(x=\dfrac{1}{35}+\dfrac{2}{7}\)
\(x=\dfrac{1}{35}+\dfrac{10}{35}\)
Vậy \(x=\dfrac{11}{35}\)
10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)
\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)
\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)
Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)

Ta có:
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
= 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= (36 – 36) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
= 0
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= (27 – 27) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7
=0
a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7
= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7
b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0
c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)
= 0 : (3 x 5 x 7 x 9 x 11)
d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7
= 0 : 9 x 1 x 3 x 5 x 7 Nếu đúng thì k cho mình nhé bạn!
Phương trình:
\(\frac{1}{9} + \left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{5}{6}\)
Bước 1: Trừ \(\frac{1}{9}\) hai vế:
\(\left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{5}{6} - \frac{1}{9}\)
Quy đồng mẫu:
Vậy:
\(\left(\left(\right. x - \frac{1}{3} \left.\right)\right)^{2} = \frac{15}{18} - \frac{2}{18} = \frac{13}{18}\)
Bước 2: Lấy căn hai vế:
\(x - \frac{1}{3} = \pm \sqrt{\frac{13}{18}}\)
Bước 3: Giải ra \(x\):
\(x = \frac{1}{3} \pm \sqrt{\frac{13}{18}}\)
Kết quả:
Đây là nghiệm dưới dạng căn thức. Nếu bạn muốn gần đúng:
Vậy:
sao em có toán 6 ?🤔