giúp với
3x - 74 = 25
x - {14 - 8 } = 4 - {23 - 3}
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\(\text{Cx khá đơn giản nhưng a giải chi tiết luôn nha}\)
\(1.\)
\(5\left(x+4\right)=25x\)
\(\Leftrightarrow5x+20=25x\)
\(\Leftrightarrow5x-25x=-20\)\(\text{(chuyển vế)}\)
\(\Leftrightarrow-20x=-20\)
\(\Leftrightarrow x=1\)
\(2.\)
\(-\frac{3}{6}=\frac{25}{2x-1}\)
\(\Leftrightarrow-\frac{1}{2}=25.\frac{1}{2x-1}\)
\(\Leftrightarrow\frac{1}{2x-1}=-\frac{1}{2}:25=-\frac{1}{50}\)
\(3.\)\(\text{(tương tự)}\)
\(\Leftrightarrow-50=2x-1\)\(\text{(nhân chéo)}\)
\(\Leftrightarrow x=-50+1=-49\)
\(4.\)
\(5\left(x-4\right)=3x+8\)
\(\Leftrightarrow5x-20=3x+8\)
\(\Leftrightarrow5x-3x=20+8\)
\(\Leftrightarrow2x=28\)
\(\Leftrightarrow x=28:2=14\)
1) 5.(x+4)=25x
5x+20=25x
20=25x-5x
20=20x
x=1
2) \(\frac{-3}{6}=\frac{25}{2x-1}\)
-3.(2x-1)=6.25
-6x+3=150
-6x=147
x=-24,5
câu 3,4 lm tương tự nha Nhớ đấy!!!!
a) \(\sqrt{x^4}=2\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x^2\right)^2}=2\)
⇔ \(\left|x^2\right|=2\)
⇔ \(\orbr{\begin{cases}x^2=2\\x^2=-2\left(loai\right)\end{cases}}\)
⇔ x2 - 2 = 0
⇔ ( x - √2 )( x + √2 ) = 0
⇔ x - √2 = 0 hoặc x + √2 = 0
⇔ x = ±√2
b) \(3\sqrt{x+1}-8=0\)( ĐK x ≥ -1 )
⇔ \(3\sqrt{x+1}=8\)
⇔ \(\sqrt{x+1}=\frac{8}{3}\)
⇔ \(x+1=\frac{64}{9}\)
⇔ \(x=\frac{55}{9}\)( tm )
c) \(2\sqrt{x-3}+\sqrt{25x-75}=14\)( ĐK x ≥ 3 ) ( Vầy hợp lí hơn á )
⇔ \(2\sqrt{x-3}+\sqrt{5^2\left(x-3\right)}=14\)
⇔ \(2\sqrt{x-3}+5\sqrt{x-3}=14\)
⇔ \(7\sqrt{x-3}=14\)
⇔ \(\sqrt{x-3}=2\)
⇔ \(x-3=4\)
⇔ \(x=7\)( tm )
d) \(\sqrt{\left(3x-1\right)^2}=5\)( ĐK x ∈ R )
⇔ \(\left|3x-1\right|=5\)
⇔ \(\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
e) \(\sqrt{x^2+4x+4}-6=0\)( ĐK x ∈ R )
⇔ \(\sqrt{\left(x+2\right)^2}=6\)
⇔ \(\left|x+2\right|=6\)
⇔ \(\orbr{\begin{cases}x+2=6\\x+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-8\end{cases}}\)
\(a)\)\(\sqrt{x^4}=2\)\(\Leftrightarrow\)\(x^2=2\)\(\Rightarrow\)\(\orbr{\begin{cases}x=\sqrt{2}\\x=-\sqrt{2}\end{cases}}\)
Vậy \(x=\sqrt{2}\)\(hoặc\)\(x=-\sqrt{2}\)
\(b)\)\(ĐK:x\ge0\)
\(3\sqrt{x+1}-8=0\)\(\Leftrightarrow\)\(3\sqrt{x}=8\)\(\Leftrightarrow\)\(\sqrt{x}=\frac{8}{3}\)\(\Leftrightarrow\)\(x=(\frac{8}{3})^2\)\(\Leftrightarrow\)\(x=\frac{64}{9}\)\((TM)\)
Vậy \(x=\frac{64}{9}\)
\(d)\)\(\sqrt{(3x-1)^2}=5\)\(\Leftrightarrow\)\(|3x-1|=5\)\((1)\)
Vậy \(x\in\hept{2;\frac{-4}{3}}\)
-Nếu \(x\ge-2\)thì \(\left(2\right)\Leftrightarrow x+2=6\Leftrightarrow x=4(TM)\)
-Nếu \(x< -2\)thì \(\left(2\right)\Leftrightarrow-\left(x+2\right)=6\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(TM\right)\)
Vậy \(x=4;x=-8\)
y: Ta có: \(x^2-x-6=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
z: Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-1\end{matrix}\right.\)
j: Ta có: \(25x^2-4=0\)
\(\Leftrightarrow\left(5x-2\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Câu 1:
\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)
\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)
\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)
\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)
\(\Leftrightarrow50x-16=0\)
\(\Leftrightarrow50x=16\)
\(\Leftrightarrow x=\dfrac{8}{25}\)
Câu 2 :
\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)
<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)
<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)
<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x
<=> 11x+27 = 26x -5
<=> ( 26x - 5 ) - ( 11x + 27 ) = 0
<=> 15x - 32 = 0
<=> 15x = 32
<=> x = \(\dfrac{32}{15}\)
a) (x^4)^2=x^12/x^5 (x khác 0 )
b) x^10+=25x^8
c) (2x+3)^2=9/121
d) (3x-1)^3=-8/27
Các bạn giúp mình với
a: \(A=4\cdot\dfrac{5}{2}\sqrt{x}-\dfrac{8}{3}\cdot\dfrac{3}{2}\sqrt{x}-\dfrac{4}{3x}\cdot\dfrac{3x}{8}\cdot\sqrt{x}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
b: \(B=\dfrac{y}{2}+\dfrac{3}{4}\cdot\left|2y-1\right|-\dfrac{3}{2}\)
\(=\dfrac{y}{2}+\dfrac{3}{4}\left(1-2y\right)-\dfrac{3}{2}\)
=1/2y+3/4-3/2y-3/2
=-y-3/4
3x - 74 = 25
3x = 25 + 74
3x = 99
x = 99 : 3
x = 33
x - {14 - 8 } = 4 - {23 - 3}
x - 6 = 4 - 20
x - 6 = -16
x = (-16) + 6
x = -10
mấy kiểu pt này bạn tải photomath về iphone nó sx giúp bạn