Giải phương trình:
a) (x-3)^2=11+6√2
b) x^2 -10x + 25=27-10√2
c) 4x^2 + 4x= 27-10√3
d) x^2+2√5x=16-4√5
e) x^2 + 4√3x=1-4√3
f) 4x^2-12√2 x -33+10√2=0
g) 2x^2-12x+9+4√2=0
h) 3x^2 -30x+26+8√3=0
Các bn giúp mk vs. Tks
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\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
a/ \(\left(x-2\right)^2=11+6\sqrt{2}\)
\(\Leftrightarrow\left(x-2\right)^2=\left(3+\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3+\sqrt{2}\\x-2=-3-\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5+\sqrt{2}\\x=-1-\sqrt{2}\end{matrix}\right.\)
b/ \(x^2-10x+25=27-10\sqrt{2}\)
\(\Leftrightarrow\left(x-5\right)^2=\left(5-\sqrt{2}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=5-\sqrt{2}\\x-5=\sqrt{2}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=10-\sqrt{2}\\x=\sqrt{2}\end{matrix}\right.\)
c/ \(4x^2+4x+1=28-10\sqrt{3}\)
\(\Leftrightarrow\left(2x+1\right)^2=\left(5-\sqrt{3}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=5-\sqrt{3}\\2x+1=\sqrt{3}-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{4-\sqrt{3}}{2}\\x=\frac{-6+\sqrt{3}}{2}\end{matrix}\right.\)
d/ \(x^2+2\sqrt{5}x+5=21-4\sqrt{5}\)
\(\Leftrightarrow\left(x+\sqrt{5}\right)^2=\left(2\sqrt{5}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{5}=2\sqrt{5}-1\\x+\sqrt{5}=1-2\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}-1\\x=1-3\sqrt{5}\end{matrix}\right.\)
e/ \(x^2+2\sqrt{12}x+12=13-4\sqrt{3}\)
\(\Leftrightarrow\left(x+2\sqrt{3}\right)^2=\left(2\sqrt{3}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2\sqrt{3}=2\sqrt{3}-1\\x+2\sqrt{3}=1-2\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1-4\sqrt{3}\end{matrix}\right.\)
f/ \(4x^2-12\sqrt{2}x+18=51-10\sqrt{2}\)
\(\Leftrightarrow\left(2x-3\sqrt{2}\right)^2=\left(5\sqrt{2}-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5\sqrt{2}=5\sqrt{2}-1\\2x-2\sqrt{2}=1-5\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{10\sqrt{2}-1}{2}\\x=\frac{1-3\sqrt{2}}{2}\end{matrix}\right.\)