a)x:3/8 + 5/8 =x
b) (5x – 3)2 – 1/64 = 0
Giải hộ mình vs ạ . Cần gấp lắm
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rồi cộng tích lại với nhau
rồi tìm x
nha bn
a) 5.3x = 405
3x = 81 = 34
=> x = 4
b) 2x : 8 = 4
2x = 32 = 25
=> x = 5
c) x64 = x
=> x64 - x = 0
x.(x63 -1) = 0
=> x = 0
x63 - 1 = 0 => x63 = 1 => x = 1
KL: ...
phần d lm tương tự phần c nha bn
e) (x-2)5 = 243 = 35
=> x - 2 = 3
x = 5
g) (x+1)3 = 125 = 53
=> x + 1 = 5
x = 4
a) \(3\left(x-5\right)+6=36\)
\(\Rightarrow3\cdot\left(x-5\right)=36-6\)
\(\Rightarrow3\cdot\left(x-5\right)=30\)
\(\Rightarrow x-5=10\)
\(\Rightarrow x=10+5\)
\(\Rightarrow x=15\)
b) \(200-8\cdot\left(x+7\right)=24\)
\(\Rightarrow8\cdot\left(x+7\right)=200-24\)
\(\Rightarrow8\cdot\left(x+7\right)=176\)
\(\Rightarrow x+7=\dfrac{176}{8}\)
\(\Rightarrow x+7=22\)
\(\Rightarrow x=22-7\)
\(\Rightarrow x=15\)
c) \(\left(x-890\right)\left(74-x\right)=0\)
+) \(x-890=0\)
\(\Rightarrow x=890\)
+) \(74-x=0\)
\(\Rightarrow x=74\)
d) \(\left(31-x\right)\cdot\left(x-64\right)=0\)
+) \(31-x=0\)
\(\Rightarrow x=31\)
+) \(x-64=0\)
\(\Rightarrow x=64\)
a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30
⇔4x+8 - 14x + 7 + 27x - 36 = 30
⇔ 17x = 51
⇔ x = 3
b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11
⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11
⇔ -14x = -4
⇔ x= \(\frac{2}{7}\)
c) 5x(1 - 2x) - 3x(x + 18) = 0
⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0
⇔ -13x\(^2\) -49 x = 0
⇔ -x ( 13x + 49 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182
⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182
⇔ 5x - 3 ( 26x - 12 ) = 182
⇔ 5x - 78x + 36 = 182
⇔ - 73x = 146
⇔ x = -2
a) \(\left(2x-3\right)^2-\left(2x+5\right)^2=10\)
\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)
\(\Leftrightarrow-32x-26=0\)
\(\Leftrightarrow-32x=26\)
\(\Rightarrow x=-\frac{13}{16}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)
\(\Leftrightarrow16x^2+4x-3=0\)
\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)
c) \(\left(x+5\right)^2=45+x^2\)
\(\Leftrightarrow x^2+10x+25-x^2-45=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
d) \(\left(2x-3\right)^2-\left(2x-1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)
\(\Leftrightarrow-8x+11=0\)
\(\Leftrightarrow-8x=-11\)
\(\Rightarrow x=\frac{11}{8}\)
e) \(\left(x-1\right)^2-\left(5x-3\right)^2=0\)
\(\Leftrightarrow\left(x-1-5x+3\right)\left(x-1+5x-3\right)=0\)
\(\Leftrightarrow\left(-4x+2\right)\left(6x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)
\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)
\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)
a) \(x:\frac{3}{8}+\frac{5}{8}=x\)
\(x.\frac{8}{3}+\frac{5}{8}=x\)
\(x-x.\frac{8}{3}=\frac{5}{8}\)
\(x\left(1-\frac{8}{3}\right)=\frac{5}{8}\)
\(x.\frac{-5}{3}=\frac{5}{8}\)
\(x=\frac{5}{8}:\frac{-5}{3}\)
\(x=\frac{-3}{8}\)
b) \(\left(5x-3\right)^2-\frac{1}{64}=0\)
\(\left(5x-3\right)^2=\frac{1}{64}\)
=> \(\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=\frac{-1}{8}\end{cases}}\)
=> \(\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)
XONG