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a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
a) \(\frac{9}{20}\) c) \(\frac{-55}{4}\)
b) \(\frac{116}{75}\) d) \(\frac{-76}{45}\)
đúng hết đấy nhé mình tính kĩ lắm ko sai đâu
chúc may mắn
Bài 1
a, 23 + ( x - 32 ) = 1
x - 32 = 1 - 23 = -7
x = -7 + 32
x = 2
b, 5 . (x+7) -10 = 40
5 . (x+7) = 50
x+7 = 50 :5 =10
x = 10 - 7
x = 3
a, ta có \(x+15⋮x+5,x+5⋮x+5\)
=>x+15-(x+5)\(⋮\)x+5
=>\(x+15-x-5⋮x+5\)
=>\(10⋮x+5\)
=>\(x+5\inƯ\left(10\right)\)=> \(x+5\in\left\{\pm1,\pm2,\pm5,\pm10\right\}\)
=>x\(\in\left\{-15,-10,-7,-6,-4,-3,0,5\right\}\)Mà x thuộc N
=> \(x\in\left\{0,5\right\}\)
Phần tiếp theo tương tự nha bn
Ta có 2x+9\(⋮x+2\)
\(x+2⋮x+2\Rightarrow2\left(x+2\right)⋮x+2\)
=> 2x+9-2(x+2)\(⋮x+2\)
=> 2x+9-2x-4\(⋮x+2\)
=>5\(⋮x+2\)
=>\(x+2\inƯ\left(5\right)\Rightarrow x+2\in\left\{\pm1,\pm5\right\}\)
=>\(x\in\left\{-7,-3,-2,3\right\}\)Mà \(x\inℕ\Rightarrow x=3\)
Vậy.........
Phần sau bn lm tương tự nhé
*****Chúc bạn học giỏi*****
\(\left|-4x+1\frac{1}{3}\right|=x+2\frac{1}{7}\)
\(\left|-4x+\frac{4}{3}\right|=x+\frac{15}{7}\)
TH1: \(-4x+\frac{4}{3}=x+\frac{15}{7}\)
\(-4x-x=\frac{15}{7}-\frac{4}{3}\)
\(-5x=\frac{17}{21}\)
x = -17/105
TH2: \(-4x+\frac{4}{3}=-x-\frac{15}{7}\)
\(-4x+x=-\frac{15}{7}-\frac{4}{3}\)
\(-3x=-\frac{73}{21}\)
x = 73/63
KL:...
a) \(\left(4x-13\right)^4+4^3=145\)
\(\Rightarrow\left(4x-13\right)^4+64=145\)
\(\Rightarrow\left(4x-13\right)^4=81\)
\(\Rightarrow4x-13=\pm3\)
+) \(4x-13=3\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
+) \(4x-13=-3\)
\(\Rightarrow4x=10\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=4\) hoặc \(x=\frac{5}{2}\)
b) \(3^{x+2}-3^x=72\)
\(\Rightarrow3^x.3^2-3^x=72\)
\(\Rightarrow3^x.\left(3^2-1\right)=72\)
\(\Rightarrow3^x.8=72\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
c) \(2^{x+2}-2^{x-1}=224\)
\(\Rightarrow2^{x-1+3}-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.2^3-2^{x-1}=224\)
\(\Rightarrow2^{x-1}.\left(2^3-1\right)=224\)
\(\Rightarrow2^{x-1}.7=224\)
\(\Rightarrow2^{x-1}=32\)
\(\Rightarrow2^{x-1}=2^5\)
\(\Rightarrow x-1=5\)
\(\Rightarrow x=6\)
Vậy x = 6
1.> => x7:x3=16
=> x4=24
=> x=4
vậy x=4
2> => x10:x3=22
=> x7=22
ko có x phù hợp